Special Series — Question 9

Question 9 – Solution

Step 1: Differentiate twice.

For $|x|<1$, $(1-x)^{-1}=\sum_{n=0}^{\infty}x^n$. Two derivatives give $$\sum_{n=2}^{\infty}n(n-1)x^{n-2}=\frac{2}{(1-x)^3}.$$ Multiplying by $x^2$ produces $$\sum_{n=2}^{\infty}n(n-1)x^n=\frac{2x^2}{(1-x)^3}.$$

Step 2: Restore the missing linear part.

Because $n^2=n(n-1)+n$, add the first-derivative identity $\sum nx^n=x/(1-x)^2$: $$\sum_{n=1}^{\infty}n^2x^n =\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2} =\frac{x(1+x)}{(1-x)^3}.$$

Step 3: Evaluate at the requested point.

Since $x=1/2$ lies inside the interval of convergence, $$\sum_{n=1}^{\infty}\frac{n^2}{2^n} =\frac{(1/2)(3/2)}{(1/2)^3}=\boxed{6}.$$ Omitting the $\sum nx^n$ term would compute $n(n-1)$ rather than $n^2$.