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Sum of series with addition

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5

1

$begingroup$

I am looking at some homework where I have:

$sum_n=1^infty frac1n^2(n+1)$

How can I sum this?

I know that

$sum_n=1^infty frac1n^2 = fracpi^26$

and also that

$sum_n=1^infty frac1n(n+1)=1$

But I just can't see or find the connection

sequences-and-series

share|cite|improve this question

asked Sep 27 at 13:50

DanielDaniel

60711 silver badge1010 bronze badges

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add a comment
 | 

5

1

$begingroup$

I am looking at some homework where I have:

$sum_n=1^infty frac1n^2(n+1)$

How can I sum this?

I know that

$sum_n=1^infty frac1n^2 = fracpi^26$

and also that

$sum_n=1^infty frac1n(n+1)=1$

But I just can't see or find the connection

sequences-and-series

share|cite|improve this question

asked Sep 27 at 13:50

DanielDaniel

60711 silver badge1010 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I am looking at some homework where I have:

$sum_n=1^infty frac1n^2(n+1)$

How can I sum this?

I know that

$sum_n=1^infty frac1n^2 = fracpi^26$

and also that

$sum_n=1^infty frac1n(n+1)=1$

But I just can't see or find the connection

sequences-and-series

share|cite|improve this question

asked Sep 27 at 13:50

DanielDaniel

60711 silver badge1010 bronze badges

$endgroup$

share|cite|improve this question

asked Sep 27 at 13:50

DanielDaniel

60711 silver badge1010 bronze badges

share|cite|improve this question

asked Sep 27 at 13:50

DanielDaniel

60711 silver badge1010 bronze badges

asked Sep 27 at 13:50

DanielDaniel

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asked Sep 27 at 13:50

DanielDaniel

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3 Answers
3

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8

$begingroup$

Hint:

$$dfrac1n^2(n+1)=dfracn+1-nn^2(n+1)=dfrac1n^2-dfrac1n(n+1)$$

$$dfrac1n(n+1)=dfracn+1-nn(n+1)=?$$

See Telescoping series

share|cite|improve this answer

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

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5

$begingroup$

HINT

$$frac1n^2(n+1)=frac1n^2+frac1n+1-frac1n$$

share|cite|improve this answer

answered Sep 27 at 13:55

useruser

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add a comment
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4

$begingroup$

Very simple
$dfrac1n^2(n+1) =dfrac(n+1)-nn^2(n+1) =dfrac1n^2 -dfrac1n(n+1) $

Can you take it from here?

share|cite|improve this answer

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but can it really be true that this sums up to some strange real number? I would think homework would be some friendly number
  $endgroup$
  – Daniel
  Sep 27 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont whether it is special or not but the answer to question is $0.644934067$
  $endgroup$
  – Akshaj Bansal
  Sep 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thanks, seems to be $pi^2/6-1$
  $endgroup$
  – Daniel
  Sep 27 at 14:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah sure its that only
  $endgroup$
  – Akshaj Bansal
  Sep 27 at 14:51
  

  
  
  
  

add a comment
 | 

Your Answer

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8

$begingroup$

Hint:

$$dfrac1n^2(n+1)=dfracn+1-nn^2(n+1)=dfrac1n^2-dfrac1n(n+1)$$

$$dfrac1n(n+1)=dfracn+1-nn(n+1)=?$$

See Telescoping series

share|cite|improve this answer

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

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$endgroup$

add a comment
 | 

8

$begingroup$

Hint:

$$dfrac1n^2(n+1)=dfracn+1-nn^2(n+1)=dfrac1n^2-dfrac1n(n+1)$$

$$dfrac1n(n+1)=dfracn+1-nn(n+1)=?$$

See Telescoping series

share|cite|improve this answer

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

243k1515 gold badges169169 silver badges292292 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Hint:

$$dfrac1n^2(n+1)=dfracn+1-nn^2(n+1)=dfrac1n^2-dfrac1n(n+1)$$

$$dfrac1n(n+1)=dfracn+1-nn(n+1)=?$$

See Telescoping series

share|cite|improve this answer

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

243k1515 gold badges169169 silver badges292292 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

243k1515 gold badges169169 silver badges292292 bronze badges

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

243k1515 gold badges169169 silver badges292292 bronze badges

answered Sep 27 at 13:55

lab bhattacharjeelab bhattacharjee

243k1515 gold badges169169 silver badges292292 bronze badges

5

$begingroup$

HINT

$$frac1n^2(n+1)=frac1n^2+frac1n+1-frac1n$$

share|cite|improve this answer

answered Sep 27 at 13:55

useruser

111k1010 gold badges4949 silver badges102102 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

HINT

$$frac1n^2(n+1)=frac1n^2+frac1n+1-frac1n$$

share|cite|improve this answer

answered Sep 27 at 13:55

useruser

111k1010 gold badges4949 silver badges102102 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

HINT

$$frac1n^2(n+1)=frac1n^2+frac1n+1-frac1n$$

share|cite|improve this answer

answered Sep 27 at 13:55

useruser

111k1010 gold badges4949 silver badges102102 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 27 at 13:55

useruser

111k1010 gold badges4949 silver badges102102 bronze badges

answered Sep 27 at 13:55

useruser

111k1010 gold badges4949 silver badges102102 bronze badges

answered Sep 27 at 13:55

useruser

111k1010 gold badges4949 silver badges102102 bronze badges

4

$begingroup$

Very simple
$dfrac1n^2(n+1) =dfrac(n+1)-nn^2(n+1) =dfrac1n^2 -dfrac1n(n+1) $

Can you take it from here?

share|cite|improve this answer

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but can it really be true that this sums up to some strange real number? I would think homework would be some friendly number
  $endgroup$
  – Daniel
  Sep 27 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont whether it is special or not but the answer to question is $0.644934067$
  $endgroup$
  – Akshaj Bansal
  Sep 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thanks, seems to be $pi^2/6-1$
  $endgroup$
  – Daniel
  Sep 27 at 14:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah sure its that only
  $endgroup$
  – Akshaj Bansal
  Sep 27 at 14:51
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Very simple
$dfrac1n^2(n+1) =dfrac(n+1)-nn^2(n+1) =dfrac1n^2 -dfrac1n(n+1) $

Can you take it from here?

share|cite|improve this answer

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but can it really be true that this sums up to some strange real number? I would think homework would be some friendly number
  $endgroup$
  – Daniel
  Sep 27 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont whether it is special or not but the answer to question is $0.644934067$
  $endgroup$
  – Akshaj Bansal
  Sep 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thanks, seems to be $pi^2/6-1$
  $endgroup$
  – Daniel
  Sep 27 at 14:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah sure its that only
  $endgroup$
  – Akshaj Bansal
  Sep 27 at 14:51
  

  
  
  
  

add a comment
 | 

$begingroup$

Very simple
$dfrac1n^2(n+1) =dfrac(n+1)-nn^2(n+1) =dfrac1n^2 -dfrac1n(n+1) $

Can you take it from here?

share|cite|improve this answer

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges

answered Sep 27 at 13:56

Akshaj BansalAkshaj Bansal

73211 silver badge88 bronze badges