What is the value of $frac11+frac13-frac15-frac17+frac19+frac111-dots$?Find the sum of $sum 1/(k^2 - a^2)$ when $0<a<1$What is the value of $sum_n=1^infty frac1n^1+varepsilon$?Calculating the infinite series $1-frac13+frac15-frac17+frac19-frac111cdots$Another SummationInfinite summation convergenceElementary way to calculate the series $sumlimits_n=1^inftyfracH_nn2^n$Convergence of $1+frac13-frac12+frac15+frac17-frac14+frac19+frac111-frac16+ldots$What is the exact value of $sumlimits_x=1^∞frac1x^x$?
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What is the value of $frac11+frac13-frac15-frac17+frac19+frac111-dots$?
Find the sum of $sum 1/(k^2 - a^2)$ when $0<a<1$What is the value of $sum_n=1^infty frac1n^1+varepsilon$?Calculating the infinite series $1-frac13+frac15-frac17+frac19-frac111cdots$Another SummationInfinite summation convergenceElementary way to calculate the series $sumlimits_n=1^inftyfracH_nn2^n$Convergence of $1+frac13-frac12+frac15+frac17-frac14+frac19+frac111-frac16+ldots$What is the exact value of $sumlimits_x=1^∞frac1x^x$?
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$begingroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $displaystylesum_k=1^infty (-1)^left(k^2 + k + 2right)/2 over 2k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
edited Apr 26 at 6:11
Felix Marin
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asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
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$begingroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $displaystylesum_k=1^infty (-1)^left(k^2 + k + 2right)/2 over 2k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
$endgroup$
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
add a comment
|
$begingroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $displaystylesum_k=1^infty (-1)^left(k^2 + k + 2right)/2 over 2k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
$endgroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $displaystylesum_k=1^infty (-1)^left(k^2 + k + 2right)/2 over 2k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
sequences-and-series summation integers pi
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
edited Apr 26 at 6:11
Felix Marin
71.2k8 gold badges116 silver badges155 bronze badges
71.2k8 gold badges116 silver badges155 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
1,2411 silver badge14 bronze badges
1,2411 silver badge14 bronze badges
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
add a comment
|
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
1
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
add a comment
|
4 Answers
4
active
oldest
votes
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
$endgroup$
4
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment
|
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
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$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
$endgroup$
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment
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$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots equiv
sum_n = 0^inftypars-1^n
sum_k = 2n + 1^2n + 21 over 2k - 1
\[5mm] = &
sum_n = 0^inftypars-1^n
sum_k = 0^11 over 2k + 4n + 1 =
sum_k = 0^1sum_n = 0^infty
pars-1^n over 4n + 2k + 1
\[5mm] = &
sum_k = 0^1sum_n = 0^inftypars-1^n
int_0^1t^4n + 2k,dd t =
sum_k = 0^1int_0^1t^2k
sum_n = 0^inftypars-t^4^n,dd t
\[5mm] = &
sum_k = 0^1int_0^1t^2k over 1 + t^4,dd t =
sum_k = 0^1int_0^1t^2k - t^2k + 4 over
1 - t^8,dd t
\[5mm] = &
1 over 8sum_k = 0^1int_0^1
t^k/4 - 7/8 - t^k/4 - 3/8 over 1 - t,dd t
\[5mm] = &
1 over 8sum_k = 0^1bracks%
Psiparsk over 4 + 5 over 8 -
Psiparsk over 4 + 1 over 8
endalign
where $dsPsi$ is the Digamma Function.
Then,
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots
\[5mm] = &
bracksPsipars5/8 - Psipars1/8 +
bracksPsipars7/8 - Psipars3/8 over 8
\[5mm] = &
bracksPsipars5/8 - Psipars3/8 +
bracksPsipars7/8 - Psipars1/8 over 8
endalign
With
Euler Reflection Formula:
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots =
picotpars3pi/8 + picotparspi/8 over 8
\[5mm] = &
pi,tanparspi/8 + cotparspi/8 over 8 =
pi over 8sinparspi/8cosparspi/8 =
pi over 4sinparspi/4
\[5mm] = &
pi over 4parsroot2/2 =
bbxroot2 over 4,pi approx 1.1107
endalign
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
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4 Answers
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oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
$endgroup$
4
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment
|
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
$endgroup$
4
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment
|
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
$endgroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
answered Apr 17 at 11:03
saulspatzsaulspatz
24.6k4 gold badges16 silver badges41 bronze badges
24.6k4 gold badges16 silver badges41 bronze badges
4
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment
|
4
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
4
4
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment
|
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
$endgroup$
add a comment
|
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
$endgroup$
add a comment
|
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
$endgroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
edited Apr 17 at 14:52
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
answered Apr 17 at 12:30
E.H.EE.H.E
18.6k1 gold badge20 silver badges70 bronze badges
18.6k1 gold badge20 silver badges70 bronze badges
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$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
$endgroup$
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment
|
$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
$endgroup$
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment
|
$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
$endgroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
edited Apr 17 at 17:49
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
answered Apr 17 at 17:33
robjohn♦robjohn
280k29 gold badges331 silver badges661 bronze badges
280k29 gold badges331 silver badges661 bronze badges
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment
|
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment
|
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots equiv
sum_n = 0^inftypars-1^n
sum_k = 2n + 1^2n + 21 over 2k - 1
\[5mm] = &
sum_n = 0^inftypars-1^n
sum_k = 0^11 over 2k + 4n + 1 =
sum_k = 0^1sum_n = 0^infty
pars-1^n over 4n + 2k + 1
\[5mm] = &
sum_k = 0^1sum_n = 0^inftypars-1^n
int_0^1t^4n + 2k,dd t =
sum_k = 0^1int_0^1t^2k
sum_n = 0^inftypars-t^4^n,dd t
\[5mm] = &
sum_k = 0^1int_0^1t^2k over 1 + t^4,dd t =
sum_k = 0^1int_0^1t^2k - t^2k + 4 over
1 - t^8,dd t
\[5mm] = &
1 over 8sum_k = 0^1int_0^1
t^k/4 - 7/8 - t^k/4 - 3/8 over 1 - t,dd t
\[5mm] = &
1 over 8sum_k = 0^1bracks%
Psiparsk over 4 + 5 over 8 -
Psiparsk over 4 + 1 over 8
endalign
where $dsPsi$ is the Digamma Function.
Then,
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots
\[5mm] = &
bracksPsipars5/8 - Psipars1/8 +
bracksPsipars7/8 - Psipars3/8 over 8
\[5mm] = &
bracksPsipars5/8 - Psipars3/8 +
bracksPsipars7/8 - Psipars1/8 over 8
endalign
With
Euler Reflection Formula:
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots =
picotpars3pi/8 + picotparspi/8 over 8
\[5mm] = &
pi,tanparspi/8 + cotparspi/8 over 8 =
pi over 8sinparspi/8cosparspi/8 =
pi over 4sinparspi/4
\[5mm] = &
pi over 4parsroot2/2 =
bbxroot2 over 4,pi approx 1.1107
endalign
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
$endgroup$
add a comment
|
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots equiv
sum_n = 0^inftypars-1^n
sum_k = 2n + 1^2n + 21 over 2k - 1
\[5mm] = &
sum_n = 0^inftypars-1^n
sum_k = 0^11 over 2k + 4n + 1 =
sum_k = 0^1sum_n = 0^infty
pars-1^n over 4n + 2k + 1
\[5mm] = &
sum_k = 0^1sum_n = 0^inftypars-1^n
int_0^1t^4n + 2k,dd t =
sum_k = 0^1int_0^1t^2k
sum_n = 0^inftypars-t^4^n,dd t
\[5mm] = &
sum_k = 0^1int_0^1t^2k over 1 + t^4,dd t =
sum_k = 0^1int_0^1t^2k - t^2k + 4 over
1 - t^8,dd t
\[5mm] = &
1 over 8sum_k = 0^1int_0^1
t^k/4 - 7/8 - t^k/4 - 3/8 over 1 - t,dd t
\[5mm] = &
1 over 8sum_k = 0^1bracks%
Psiparsk over 4 + 5 over 8 -
Psiparsk over 4 + 1 over 8
endalign
where $dsPsi$ is the Digamma Function.
Then,
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots
\[5mm] = &
bracksPsipars5/8 - Psipars1/8 +
bracksPsipars7/8 - Psipars3/8 over 8
\[5mm] = &
bracksPsipars5/8 - Psipars3/8 +
bracksPsipars7/8 - Psipars1/8 over 8
endalign
With
Euler Reflection Formula:
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots =
picotpars3pi/8 + picotparspi/8 over 8
\[5mm] = &
pi,tanparspi/8 + cotparspi/8 over 8 =
pi over 8sinparspi/8cosparspi/8 =
pi over 4sinparspi/4
\[5mm] = &
pi over 4parsroot2/2 =
bbxroot2 over 4,pi approx 1.1107
endalign
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
$endgroup$
add a comment
|
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots equiv
sum_n = 0^inftypars-1^n
sum_k = 2n + 1^2n + 21 over 2k - 1
\[5mm] = &
sum_n = 0^inftypars-1^n
sum_k = 0^11 over 2k + 4n + 1 =
sum_k = 0^1sum_n = 0^infty
pars-1^n over 4n + 2k + 1
\[5mm] = &
sum_k = 0^1sum_n = 0^inftypars-1^n
int_0^1t^4n + 2k,dd t =
sum_k = 0^1int_0^1t^2k
sum_n = 0^inftypars-t^4^n,dd t
\[5mm] = &
sum_k = 0^1int_0^1t^2k over 1 + t^4,dd t =
sum_k = 0^1int_0^1t^2k - t^2k + 4 over
1 - t^8,dd t
\[5mm] = &
1 over 8sum_k = 0^1int_0^1
t^k/4 - 7/8 - t^k/4 - 3/8 over 1 - t,dd t
\[5mm] = &
1 over 8sum_k = 0^1bracks%
Psiparsk over 4 + 5 over 8 -
Psiparsk over 4 + 1 over 8
endalign
where $dsPsi$ is the Digamma Function.
Then,
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots
\[5mm] = &
bracksPsipars5/8 - Psipars1/8 +
bracksPsipars7/8 - Psipars3/8 over 8
\[5mm] = &
bracksPsipars5/8 - Psipars3/8 +
bracksPsipars7/8 - Psipars1/8 over 8
endalign
With
Euler Reflection Formula:
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots =
picotpars3pi/8 + picotparspi/8 over 8
\[5mm] = &
pi,tanparspi/8 + cotparspi/8 over 8 =
pi over 8sinparspi/8cosparspi/8 =
pi over 4sinparspi/4
\[5mm] = &
pi over 4parsroot2/2 =
bbxroot2 over 4,pi approx 1.1107
endalign
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
$endgroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots equiv
sum_n = 0^inftypars-1^n
sum_k = 2n + 1^2n + 21 over 2k - 1
\[5mm] = &
sum_n = 0^inftypars-1^n
sum_k = 0^11 over 2k + 4n + 1 =
sum_k = 0^1sum_n = 0^infty
pars-1^n over 4n + 2k + 1
\[5mm] = &
sum_k = 0^1sum_n = 0^inftypars-1^n
int_0^1t^4n + 2k,dd t =
sum_k = 0^1int_0^1t^2k
sum_n = 0^inftypars-t^4^n,dd t
\[5mm] = &
sum_k = 0^1int_0^1t^2k over 1 + t^4,dd t =
sum_k = 0^1int_0^1t^2k - t^2k + 4 over
1 - t^8,dd t
\[5mm] = &
1 over 8sum_k = 0^1int_0^1
t^k/4 - 7/8 - t^k/4 - 3/8 over 1 - t,dd t
\[5mm] = &
1 over 8sum_k = 0^1bracks%
Psiparsk over 4 + 5 over 8 -
Psiparsk over 4 + 1 over 8
endalign
where $dsPsi$ is the Digamma Function.
Then,
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots
\[5mm] = &
bracksPsipars5/8 - Psipars1/8 +
bracksPsipars7/8 - Psipars3/8 over 8
\[5mm] = &
bracksPsipars5/8 - Psipars3/8 +
bracksPsipars7/8 - Psipars1/8 over 8
endalign
With
Euler Reflection Formula:
beginalign
&bbox[10px,#ffd]1 over 1 + 1 over 3 - 1 over 5
- 1 over 7 + 1 over 9 + 1 over 11 - cdots =
picotpars3pi/8 + picotparspi/8 over 8
\[5mm] = &
pi,tanparspi/8 + cotparspi/8 over 8 =
pi over 8sinparspi/8cosparspi/8 =
pi over 4sinparspi/4
\[5mm] = &
pi over 4parsroot2/2 =
bbxroot2 over 4,pi approx 1.1107
endalign
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
edited Apr 26 at 7:41
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
answered Apr 26 at 7:01
Felix MarinFelix Marin
71.2k8 gold badges116 silver badges155 bronze badges
71.2k8 gold badges116 silver badges155 bronze badges
add a comment
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add a comment
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1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40