Determine whether or not $sum_k=1^inftyleft(frac kk+1right)^k^2$ converges.Determine whether the series $sum_n=1^infty left ( fracpi2-arctan n right )$ converges or not.How to determine whether $sum_n=1^inftylnleft(fracn+2n+1right)$ converges or diverges.Determine whether the series $sum_n=1^+inftyleft(1+frac1nright)a_n$ is convergent or divergentConvergence for $sum _n=1^infty :fracsqrt[4]n^2-1sqrtn^4-1$Converge? $sum_k=1^inftyfrac sin left(frac1kright) k $Decide whether the series $sum_n=1^infty frac1+5^n1+6^n$ converges or divergesDetermine whether the series converges or diverges.To test whether $sum_n=1^inftyfracn+22^n+3sinleft[(n+frac12)piright]$ convergesDetermining whether the series: $sum_n=1^infty tanleft(frac1nright) $ convergesDetermine the convergence/divergence of $sum_n=1^inftyfraclnn!n^3$
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Determine whether or not $sum_k=1^inftyleft(frac kk+1right)^k^2$ converges.
Determine whether the series $sum_n=1^infty left ( fracpi2-arctan n right )$ converges or not.How to determine whether $sum_n=1^inftylnleft(fracn+2n+1right)$ converges or diverges.Determine whether the series $sum_n=1^+inftyleft(1+frac1nright)a_n$ is convergent or divergentConvergence for $sum _n=1^infty :fracsqrt[4]n^2-1sqrtn^4-1$Converge? $sum_k=1^inftyfrac sin left(frac1kright) k $Decide whether the series $sum_n=1^infty frac1+5^n1+6^n$ converges or divergesDetermine whether the series converges or diverges.To test whether $sum_n=1^inftyfracn+22^n+3sinleft[(n+frac12)piright]$ convergesDetermining whether the series: $sum_n=1^infty tanleft(frac1nright) $ convergesDetermine the convergence/divergence of $sum_n=1^inftyfraclnn!n^3$
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$$sum_k=1^inftyleft(frac kk+1right)^k^2$$
Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.
sequences-and-series convergence
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
$endgroup$
add a comment |
$begingroup$
$$sum_k=1^inftyleft(frac kk+1right)^k^2$$
Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.
sequences-and-series convergence
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
$endgroup$
$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38
$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40
1
$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44
1
$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14
add a comment |
$begingroup$
$$sum_k=1^inftyleft(frac kk+1right)^k^2$$
Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.
sequences-and-series convergence
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
$endgroup$
$$sum_k=1^inftyleft(frac kk+1right)^k^2$$
Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.
sequences-and-series convergence
sequences-and-series convergence
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
edited Apr 15 at 6:19
user21820
42k5 gold badges48 silver badges170 bronze badges
42k5 gold badges48 silver badges170 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
asked Apr 15 at 2:33
radiusradius
612 bronze badges
612 bronze badges
$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38
$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40
1
$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44
1
$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14
add a comment |
$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38
$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40
1
$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44
1
$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14
$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38
$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38
$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40
$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40
1
1
$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44
$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44
1
1
$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14
$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$
So your series converges!
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
$endgroup$
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
1
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
add a comment |
$begingroup$
Hint: $$left( frackk+1 right)^k sim e^-1 $$
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
$endgroup$
add a comment |
$begingroup$
$$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$
$$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
$$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
$$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$
So your series converges!
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
$endgroup$
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
1
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
add a comment |
$begingroup$
By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$
So your series converges!
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
$endgroup$
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
1
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
add a comment |
$begingroup$
By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$
So your series converges!
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
$endgroup$
By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$
So your series converges!
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
edited Apr 15 at 2:53
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
answered Apr 15 at 2:46
Chinnapparaj RChinnapparaj R
8,7612 gold badges10 silver badges32 bronze badges
8,7612 gold badges10 silver badges32 bronze badges
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
1
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
add a comment |
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
1
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
$begingroup$
How did you know to take the supremum
$endgroup$
– radius
Apr 15 at 2:51
1
1
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
$begingroup$
See the wiki link!
$endgroup$
– Chinnapparaj R
Apr 15 at 2:54
add a comment |
$begingroup$
Hint: $$left( frackk+1 right)^k sim e^-1 $$
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
$endgroup$
add a comment |
$begingroup$
Hint: $$left( frackk+1 right)^k sim e^-1 $$
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
$endgroup$
add a comment |
$begingroup$
Hint: $$left( frackk+1 right)^k sim e^-1 $$
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
$endgroup$
Hint: $$left( frackk+1 right)^k sim e^-1 $$
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
answered Apr 15 at 2:38
Robert IsraelRobert Israel
345k23 gold badges241 silver badges503 bronze badges
345k23 gold badges241 silver badges503 bronze badges
add a comment |
add a comment |
$begingroup$
$$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$
$$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
$$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
$$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
$endgroup$
add a comment |
$begingroup$
$$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$
$$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
$$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
$$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
$endgroup$
add a comment |
$begingroup$
$$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$
$$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
$$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
$$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
$endgroup$
$$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$
$$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
$$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
$$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
answered Apr 15 at 5:25
Claude LeiboviciClaude Leibovici
133k11 gold badges61 silver badges142 bronze badges
133k11 gold badges61 silver badges142 bronze badges
add a comment |
add a comment |
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$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38
$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40
1
$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44
1
$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14