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$x$
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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$x$
$begingroup$
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
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.
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
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/…
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– 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
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.
$endgroup$
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.
This question already has an answer here:
How to prove $sum_n=0^infty fracn^22^n = 6$?
6 answers
sequences-and-series power-series
sequences-and-series power-series
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
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
edited May 14 at 23:24
David G. Stork
12.7k41837
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
asked May 14 at 23:21
Erinç Emre ÇeliktenErinç Emre Çelikten
183
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.
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.
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.
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
$begingroup$
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
3
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
$begingroup$
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
$begingroup$
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
$begingroup$
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 Answers
3
active
oldest
votes
$begingroup$
$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
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 $x=frac13$
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 |
$begingroup$
Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*
answered May 14 at 23:30
Donald SplutterwitDonald Splutterwit
23.5k21448
$endgroup$
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.
answered May 14 at 23:37
Feng ShaoFeng Shao
19911
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
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 $x=frac13$
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 $x=frac13$
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 $x=frac13$
edited May 14 at 23:33
clathratus
5,7051443
answered May 14 at 23:29
E.H.EE.H.E
17.9k11969
$endgroup$
$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
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 $x=frac13$
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
edited May 14 at 23:33
clathratus
5,7051443
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
answered May 14 at 23:29
E.H.EE.H.E
17.9k11969
17.9k11969
add a comment |
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*
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*
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*
answered May 14 at 23:30
Donald SplutterwitDonald Splutterwit
23.5k21448
$endgroup$
Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*
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
answered May 14 at 23:30
Donald SplutterwitDonald Splutterwit
23.5k21448
23.5k21448
add a comment |
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.
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.
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.
answered May 14 at 23:37
Feng ShaoFeng Shao
19911
$endgroup$
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.
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
answered May 14 at 23:37
Feng ShaoFeng Shao
19911
19911
add a comment |
add a comment |
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