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Computing a series sum [duplicate]

How to prove $sum_n=0^infty fracn^22^n = 6$?Writing an infinite series as the sum of the seriesComputing the sum of an infinite seriescomputing the series $sum_n=1^infty frac1n^2 2^n$Sum of a series to an exact answerConfused computing sum of Fourier seriesClosed form of this series?Prove that the given series is convergent.Sum of reciprocals of the square roots of the first N Natural NumbersSum function for power seriesShowing the sum of a power series is less than P$x$

3

$begingroup$

This question already has an answer here:

• 
  How to prove $sum_n=0^infty fracn^22^n = 6$?
  
  6 answers
  
  

$$sum_k=0^infty frack^23^k$$

I tried with that 2 method but I couldn't get the $n^2$ term.

sequences-and-series power-series

share|cite|improve this question

edited May 14 at 23:24

David G. Stork

12.7k41837

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

New contributor

Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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marked as duplicate by Hans Lundmark, Yanior Weg, Lord Shark the Unknown, Lee David Chung Lin, Cesareo May 16 at 6:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
  $endgroup$
  – Cameron Williams
  May 14 at 23:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/593996/…
  $endgroup$
  – lab bhattacharjee
  May 15 at 2:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax.
  $endgroup$
  – dantopa
  May 16 at 1:52
  

  
  
  
  

add a comment | 

3

$begingroup$

This question already has an answer here:

• 
  How to prove $sum_n=0^infty fracn^22^n = 6$?
  
  6 answers
  
  

$$sum_k=0^infty frack^23^k$$

I tried with that 2 method but I couldn't get the $n^2$ term.

sequences-and-series power-series

share|cite|improve this question

edited May 14 at 23:24

David G. Stork

12.7k41837

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

New contributor

Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by Hans Lundmark, Yanior Weg, Lord Shark the Unknown, Lee David Chung Lin, Cesareo May 16 at 6:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You'll need to take two derivatives here and also make use of the first derivative case since taking two derivatives gives you $k(k-1) = k^2 - k$.
  $endgroup$
  – Cameron Williams
  May 14 at 23:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/593996/…
  $endgroup$
  – lab bhattacharjee
  May 15 at 2:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax.
  $endgroup$
  – dantopa
  May 16 at 1:52
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  How to prove $sum_n=0^infty fracn^22^n = 6$?
  
  6 answers
  
  

$$sum_k=0^infty frack^23^k$$

I tried with that 2 method but I couldn't get the $n^2$ term.

sequences-and-series power-series

share|cite|improve this question

edited May 14 at 23:24

David G. Stork

12.7k41837

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

New contributor

Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited May 14 at 23:24

David G. Stork

12.7k41837

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

New contributor

Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited May 14 at 23:24

David G. Stork

12.7k41837

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

New contributor

Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited May 14 at 23:24

David G. Stork

12.7k41837

edited May 14 at 23:24

David G. Stork

12.7k41837

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

New contributor

Erinç Emre Çelikten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 14 at 23:21

Erinç Emre ÇeliktenErinç Emre Çelikten

183

3 Answers
3

active

oldest

votes

5

$begingroup$

$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
Multiply by $x$
$$fracx(1-x)^2=sum_k=1^inftykx^k$$
$$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
Multiply by $x$
$$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
then let $x=frac13$

share|cite|improve this answer

edited May 14 at 23:33

clathratus

5,7051443

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

$endgroup$

add a comment | 

0

$begingroup$

Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*

share|cite|improve this answer

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

add a comment | 

0

$begingroup$

There is another way to deal with this problem.

Denote $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.

In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.

share|cite|improve this answer

answered May 14 at 23:37

Feng ShaoFeng Shao

19911

$endgroup$

add a comment | 

5

$begingroup$

$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
Multiply by $x$
$$fracx(1-x)^2=sum_k=1^inftykx^k$$
$$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
Multiply by $x$
$$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
then let $x=frac13$

share|cite|improve this answer

edited May 14 at 23:33

clathratus

5,7051443

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

$endgroup$

add a comment | 

5

$begingroup$

$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
Multiply by $x$
$$fracx(1-x)^2=sum_k=1^inftykx^k$$
$$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
Multiply by $x$
$$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
then let $x=frac13$

share|cite|improve this answer

edited May 14 at 23:33

clathratus

5,7051443

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

$endgroup$

add a comment | 

$begingroup$

$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
Multiply by $x$
$$fracx(1-x)^2=sum_k=1^inftykx^k$$
$$left(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k-1$$
Multiply by $x$
$$xleft(fracx(1-x)^2right)'=sum_k=1^inftyk^2x^k$$
then let $x=frac13$

share|cite|improve this answer

edited May 14 at 23:33

clathratus

5,7051443

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

$endgroup$

share|cite|improve this answer

edited May 14 at 23:33

clathratus

5,7051443

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

edited May 14 at 23:33

clathratus

5,7051443

edited May 14 at 23:33

clathratus

5,7051443

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

answered May 14 at 23:29

E.H.EE.H.E

17.9k11969

0

$begingroup$

Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*

share|cite|improve this answer

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

add a comment | 

0

$begingroup$

Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*

share|cite|improve this answer

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

add a comment | 

$begingroup$

Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*

share|cite|improve this answer

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

$endgroup$

share|cite|improve this answer

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

answered May 14 at 23:30

Donald SplutterwitDonald Splutterwit

23.5k21448

0

$begingroup$

There is another way to deal with this problem.

Denote $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.

In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.

share|cite|improve this answer

answered May 14 at 23:37

Feng ShaoFeng Shao

19911

$endgroup$

add a comment | 

0

$begingroup$

There is another way to deal with this problem.

Denote $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.

In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.

share|cite|improve this answer

answered May 14 at 23:37

Feng ShaoFeng Shao

19911

$endgroup$

add a comment | 

$begingroup$

There is another way to deal with this problem.

Denote $S_n=sum_k=0^n frack^23^k$ then$frac13S_n=sum_k=1^n+1 frac(k-1)^23^k$ so $frac23S_n=sum_k=1^nfrac2k-13^k-fracn^23^n+1$.

In this way we change the numerator from twice power of $k$ to the lower power. Using the same operation you can change the numerator to numbers without the appearance of $k$ then you can use the summation formula for geometric series.

share|cite|improve this answer

answered May 14 at 23:37

Feng ShaoFeng Shao

19911

$endgroup$

share|cite|improve this answer

answered May 14 at 23:37

Feng ShaoFeng Shao

19911

answered May 14 at 23:37

Feng ShaoFeng Shao

19911

answered May 14 at 23:37

Feng ShaoFeng Shao

19911