I am getting $\sum_{n=1}^{\infty}\frac{n^{2}}{2^{n}}=\frac{1}{16}$ but it is wrong according to wolfram alpha. Where is the error in my solution?

Question

This is my solution - $$ \frac{d^{2}}{dx^{2}}\left(\frac{1}{1-x}\right)=\frac{d^{2}}{dx^{2}}\sum_{n=0}^{\infty}x^{n}=\frac{d}{dx}\sum_{n=1}^{\infty}nx^{\left(n-1\right)}=\sum_{n=1}^{\infty}n\left(n-1\right)x^{\left(n-2\right)} \\ \frac{d^{2}}{dx^{2}}\left(\frac{1}{1-x}\right) = \sum_{n=1}^{\infty}n\left(n-1\right)x^{\left(n-2\right)}$$ Since this is absolutely convergent for $|x| < 1$, it is commutative. I haven't changed the lower bound from n = 1 to n = 2 since the term at n = 1 is 0 anyways. $$ \frac{d^{2}}{dx^{2}}\left(\frac{1}{1-x}\right)=\sum_{n=1}^{\infty}n^{2}x^{\left(n-2\right)}-\sum_{n=1}^{\infty}nx^{\left(n-2\right)} \\ x^{2}\frac{d^{2}}{dx^{2}}\left(\frac{1}{1-x}\right)=\sum_{n=1}^{\infty}n^{2}x^{n}-x\sum_{n=1}^{\infty}nx^{\left(n-1\right)} \\ x^{2}\frac{d^{2}}{dx^{2}}\left(\frac{1}{1-x}\right)=\sum_{n=1}^{\infty}n^{2}x^{n}-x\frac{d}{dx}\sum_{n=1}^{\infty}x^{n}\\ x^{2}\frac{d^{2}}{dx^{2}}\left(\frac{1}{1-x}\right)=\sum_{n=1}^{\infty}n^{2}x^{n}-x\frac{d}{dx}\left(\frac{1}{1-x}\right)\\ \sum_{n=1}^{\infty}n^{2}x^{n}=x^{2}\left(\frac{2}{\left(1-x\right)^{3}}\right)+x\left(\frac{1}{\left(1-x\right)^{2}}\right)\\ \sum_{n=1}^{\infty}\frac{n^{2}}{2^{n}}=\sum_{n=1}^{\infty}n^{2}\left(\frac{1}{2}\right)^{n}=\frac{1}{2^{3}}-\frac{1}{2^{4}}=\frac{1}{2^{4}} $$

Answer 1

Notice that while the second derivative can be written as

$$\frac{d^2}{dx^2}\left[\frac{1}{1-x}\right] = \sum_{n=0}^\infty n(n-1)x^{n-2},$$

for $n \in \{0, 1\}$, the summand is zero:

$$\sum_{n=0}^\infty n(n-1)x^{n-2} = 0 + 0 + 2 + 6x + 12x^2 + \cdots.$$

Hence it is better to adjust the lower index of summation before continuing to manipulate the sum:

$$\frac{d^2}{dx^2}\left[\frac{1}{1-x}\right] = \sum_{n=\color{red}{2}}^\infty n(n-1)x^{n-2}.$$ Then continuing with your approach,

$$\begin{align} x^2 \frac{d^2}{dx^2}\left[\frac{1}{1-x}\right] &= \sum_{n=2}^\infty n^2 x^n - \sum_{n=2}^\infty nx^n \\ &= -x + \sum_{n=1}^\infty n^2 x^n + x - \sum_{n=1}^\infty nx^n \\ &= \sum_{n=1}^\infty n^2 x^n - x \sum_{n=1}^\infty nx^{n-1} \\ &= \sum_{n=1}^\infty n^2 x^n - x \frac{d}{dx}\left[\frac{1}{1-x}\right]. \end{align}$$

This implies $$\sum_{n=1}^\infty n^2 x^n = \frac{2x^2}{(1-x)^3} + \frac{x}{(1-x)^2} = \frac{x(1+x)}{(1-x)^3}.$$

However, there is a much more direct way to do the calculation:

$$\frac{d}{dx}\left[\frac{1}{1-x}\right] = \sum_{n=1}^\infty nx^{n-1}$$ implies $$x \frac{d}{dx}\left[\frac{1}{1-x}\right] = \sum_{n=1}^\infty nx^n$$ which implies $$\frac{d}{dx}\left[x \frac{d}{dx}\left[\frac{1}{1-x}\right]\right] = \sum_{n=1}^\infty n^2 x^{n-1}.$$

All that is left is to compute the LHS:

$$\frac{d}{dx}\left[x \frac{d}{dx}\left[\frac{1}{1-x}\right]\right] = \frac{d}{dx}\left[\frac{x}{(1-x)^2}\right] = \frac{1+x}{(1-x)^3}.$$ Then multiplying everything by $x$ gives the desired result:

$$\frac{x(1+x)}{(1-x)^3} = \sum_{n=1}^\infty n^2 x^n.$$

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It would seem that your computation is mostly correct, except you have made an arithmetic error in the very last line. You should have $$\left(\frac{1}{2}\right)^2 \frac{2}{(1-\frac{1}{2})^3} = \frac{1}{4} \cdot \frac{2}{(1/2)^3} = \frac{16}{4} = 4,$$ and $$\frac{1}{2} \frac{1}{(1-\frac{1}{2})^2} = \frac{1}{2} \cdot 4 = 2.$$

Answer 2

I have not looked at your arguments. The terms in your summation are all positive. Simply looking at the first 2 terms: $n =1$, gives $1/2$ and $n=2$ gives $2^2/2^2=1$ and so the sum should be $>3/2$, and can't be $1/16$.