Computing a series sumHow 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
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$x$
$begingroup$
$$sum_k=0^infty frack^23^k$$
I tried with that 2 method but I couldn't get the $n^2$ term.
sequences-and-series power-series
edited 6 hours ago
David G. Stork
12.5k41836
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
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.
$endgroup$
add a comment |
$begingroup$
$$sum_k=0^infty frack^23^k$$
I tried with that 2 method but I couldn't get the $n^2$ term.
sequences-and-series power-series
edited 6 hours ago
David G. Stork
12.5k41836
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
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.
$endgroup$
2
$begingroup$
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
6 hours ago
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
9 mins ago
add a comment |
$begingroup$
$$sum_k=0^infty frack^23^k$$
I tried with that 2 method but I couldn't get the $n^2$ term.
sequences-and-series power-series
edited 6 hours ago
David G. Stork
12.5k41836
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
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.
$endgroup$
$$sum_k=0^infty frack^23^k$$
I tried with that 2 method but I couldn't get the $n^2$ term.
sequences-and-series power-series
sequences-and-series power-series
edited 6 hours ago
David G. Stork
12.5k41836
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
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 6 hours ago
David G. Stork
12.5k41836
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
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 6 hours ago
David G. Stork
12.5k41836
edited 6 hours ago
David G. Stork
12.5k41836
edited 6 hours ago
David G. Stork
12.5k41836
12.5k41836
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
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 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
asked 6 hours ago
Erinç Emre ÇeliktenErinç Emre Çelikten
182
182
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
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Check out our Code of Conduct.
2
$begingroup$
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
6 hours ago
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
9 mins ago
add a comment |
2
$begingroup$
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
6 hours ago
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
9 mins ago
2
2
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
9 mins ago
$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
9 mins ago
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 6 hours ago
clathratus
5,6811441
answered 6 hours ago
E.H.EE.H.E
17.5k11969
$endgroup$
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 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
$endgroup$
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 6 hours ago
Feng ShaoFeng Shao
18010
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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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 6 hours ago
clathratus
5,6811441
answered 6 hours ago
E.H.EE.H.E
17.5k11969
$endgroup$
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 6 hours ago
clathratus
5,6811441
answered 6 hours ago
E.H.EE.H.E
17.5k11969
$endgroup$
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 6 hours ago
clathratus
5,6811441
answered 6 hours ago
E.H.EE.H.E
17.5k11969
$endgroup$
$$frac11-x=sum_k=0^inftyx^k$$
$$frac1(1-x)^2=sum_k=1^inftykx^k-1$$
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 6 hours ago
clathratus
5,6811441
answered 6 hours ago
E.H.EE.H.E
17.5k11969
edited 6 hours ago
clathratus
5,6811441
edited 6 hours ago
clathratus
5,6811441
edited 6 hours ago
clathratus
5,6811441
5,6811441
answered 6 hours ago
E.H.EE.H.E
17.5k11969
answered 6 hours ago
E.H.EE.H.E
17.5k11969
answered 6 hours ago
E.H.EE.H.E
17.5k11969
17.5k11969
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 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
$endgroup$
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 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
$endgroup$
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 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
$endgroup$
Hint try to show
begineqnarray*
sum_k=0^infty k^2 x^k =fracx(1+4x+x^2)(1-x)^3.
endeqnarray*
answered 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
answered 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
answered 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
answered 6 hours ago
Donald SplutterwitDonald Splutterwit
23.4k21448
23.4k21448
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 6 hours ago
Feng ShaoFeng Shao
18010
$endgroup$
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 6 hours ago
Feng ShaoFeng Shao
18010
$endgroup$
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 6 hours ago
Feng ShaoFeng Shao
18010
$endgroup$
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 6 hours ago
Feng ShaoFeng Shao
18010
answered 6 hours ago
Feng ShaoFeng Shao
18010
answered 6 hours ago
Feng ShaoFeng Shao
18010
answered 6 hours ago
Feng ShaoFeng Shao
18010
18010
add a comment |
add a comment |
Erinç Emre Çelikten is a new contributor. Be nice, and check out our Code of Conduct.
Erinç Emre Çelikten is a new contributor. Be nice, and check out our Code of Conduct.
Erinç Emre Çelikten is a new contributor. Be nice, and check out our Code of Conduct.
Erinç Emre Çelikten is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
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
6 hours ago
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Possible duplicate of How to prove $sum_n=0^infty fracn^22^n = 6$?
$endgroup$
– Hans Lundmark
9 mins ago