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Sum of series with addition

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Sum of series with addition

summation of series with residue formulaWriting an infinite series as the sum of the seriesInfinite Series with PiContinuous and differentiable sum of seriesSum of an infinite series seriesValue of $sumlimits_n=1^inftyfrac1(n+1)^2-1$Finding the sum of an alternating seriesSum of an infinite geometric series with squared powersSum to infinity seriesfinding the sum of power series

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2

$begingroup$

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 9 hours ago

DennisDennis

2177 bronze badges

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add a comment
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2

$begingroup$

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 9 hours ago

DennisDennis

2177 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 9 hours ago

DennisDennis

2177 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

DennisDennis

2177 bronze badges

share|cite|improve this question

asked 9 hours ago

DennisDennis

2177 bronze badges

asked 9 hours ago

DennisDennis

2177 bronze badges

asked 9 hours ago

DennisDennis

2177 bronze badges

3 Answers
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5

$begingroup$

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

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answered 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 bronze badges

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3

$begingroup$

HINT

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

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

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3

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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 9 hours ago

Akshaj BansalAkshaj Bansal

40488 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
  $endgroup$
  – Dennis
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont whether it is special or not but the answer to question is $0.644934067$
  $endgroup$
  – Akshaj Bansal
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thanks, seems to be $pi^2/6-1$
  $endgroup$
  – Dennis
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah sure its that only
  $endgroup$
  – Akshaj Bansal
  8 hours ago
  

  
  
  
  

add a comment
 | 

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5

$begingroup$

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 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 bronze badges

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add a comment
 | 

5

$begingroup$

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 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 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 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 bronze badges

answered 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 bronze badges

answered 9 hours ago

lab bhattacharjeelab bhattacharjee

238k1515 gold badges167167 silver badges289289 bronze badges

3

$begingroup$

HINT

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

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

HINT

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

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

HINT

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

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

answered 9 hours ago

gimusigimusi

94.4k88 gold badges4646 silver badges9595 bronze badges

3

$begingroup$

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 9 hours ago

Akshaj BansalAkshaj Bansal

40488 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
  $endgroup$
  – Dennis
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont whether it is special or not but the answer to question is $0.644934067$
  $endgroup$
  – Akshaj Bansal
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thanks, seems to be $pi^2/6-1$
  $endgroup$
  – Dennis
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah sure its that only
  $endgroup$
  – Akshaj Bansal
  8 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

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 9 hours ago

Akshaj BansalAkshaj Bansal

40488 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
  $endgroup$
  – Dennis
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont whether it is special or not but the answer to question is $0.644934067$
  $endgroup$
  – Akshaj Bansal
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok thanks, seems to be $pi^2/6-1$
  $endgroup$
  – Dennis
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah sure its that only
  $endgroup$
  – Akshaj Bansal
  8 hours ago
  

  
  
  
  

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 9 hours ago

Akshaj BansalAkshaj Bansal

40488 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Akshaj BansalAkshaj Bansal

40488 bronze badges

answered 9 hours ago

Akshaj BansalAkshaj Bansal

40488 bronze badges

answered 9 hours ago

Akshaj BansalAkshaj Bansal

40488 bronze badges