Find the sum of $$\sum_{n=1}^{\infty} \frac{n^2}{2^n}$$ I know I need to manipulate the power series $\sum_{n=0}^{\infty}x^n$ with $x = \frac{1}{2}$, but I'm not sure how. Would differentiating it twice help?
Asked
Active
Viewed 167 times
1
-
3http://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 – lab bhattacharjee Nov 22 '16 at 03:57
-
Does this answer your question? How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? – Hans Lundmark May 10 '22 at 10:21
2 Answers
1
Hint
Considering $$S=\sum_{i=1}^\infty n^2 x^n=\sum_{i=1}^\infty (n^2-n+n) x^n=\sum_{i=1}^\infty n(n-1) x^n+\sum_{i=1}^\infty n x^n$$ $$S=x^2\sum_{i=1}^\infty n(n-1) x^{n-2}+x\sum_{i=1}^\infty n x^{n-1}$$ $$S=x^2\left( \sum_{i=1}^\infty x^{n}\right)''+x\left( \sum_{i=1}^\infty x^{n}\right)'$$
Claude Leibovici
- 260,315
0
HINT:
If $S(x)=\sum_{n=1}^\infty x^n=\frac{1}{1-x}$ for $|x|<1$, then
$$x(xS'(x))'=\sum_{n=1}^\infty n^2x^n$$
Then, let $x=\frac12$.
Mark Viola
- 179,405
-
Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Dec 02 '16 at 14:33