Sum of a series - 5

Question

Can anyone simplify this series,

$$S= \sum_{n=2}^{\infty} \frac{n(n-1)}{2^{2n-1}}$$ into a fraction?

Is there a closed form solution for this?

This seems to the answer for one other problem that I am working on?

Any help is appreciated.

Answer 1

HINT:

Let $\displaystyle S=\sum_{r=0}^nr(r-1)a^r=\sum_{r=2}^nr(r-1)a^r $

$\displaystyle\implies a\cdot S=\sum_{r=2}^nr(r-1)a^{r+1}=\sum_{r=3}^{n+1}(r-1)(r-2)a^r $

$\displaystyle\implies S(1-a)=2a^2+\sum_{r=3}^n2(r-1)a^r-n(n-1)a^{n+1}$ $\displaystyle S(1-a)=2a^2-n(n-1)a^{n+1}+2a\sum_{r=3}^n(r-1)a^{r-1}$

Again let $\displaystyle T=\sum_{r=3}^n(r-1)a^{r-1}$

$\displaystyle T-T\cdot a=?$

Can you recognize the Geometric Series?

Now for $|a|<1,$ use Sequence Limit: $\lim\limits_{n \rightarrow \infty}{n\,x^n}$ or How to prove that $\lim_{n \to \infty} n x^{n} = 0 $ when $0<x<1$?

Can you prove $\lim_{n\to\infty}n(n-1)a^n=0$ for $|a|<1?$

Answer 2

HINT:

Using Infinite Geometric Series $$\sum_{n=0}^\infty a^n=\frac1{1-a}$$ for $|a|<1$

Differentiate w.r.t $a$ to find $$\sum_{n=0}^\infty na^{n-1}=\frac1{(1-a)^2}$$ (See this)

Again, differentiate w.r.t $a$

Now, $$\sum_{n=2}^{\infty} \frac{n(n-1)}{2^{2n-1}}=2\sum_{n=2}^{\infty}n(n-1)\left(\frac14\right)^n$$