Ask Uncle Colin: a surprising sum

Dear Uncle Colin,

Apparently, $\sum _1^{\mathrm{\infty }}\left(\frac{n^2}{2^n}\right)=6$, which surprised me. Can you explain why?

- Some Intuition Gone Missing, Aaargh!

Hi, SIGMA, and thanks for your message! I’m not sure I can give you intuition, but I can explain where the result comes from.

Let’s start from the binomial expansion for $(1-x{)}^{-2}$, which is $1+2x+3x^2+\cdots +(n+1)x^n+\dots$.

Multiply that by $x$: $x(1-x{)}^{-2}=x+2x^2+3x^3+\cdots +n{x}^n+\dots$

Differentiate that: $(1-x{)}^{-2}+2x(1-x{)}^{-3}=1+4x+9x^2+\cdots +n^2{x}^{n-1}+\dots$

Multiply by $x$ again: $x(1-x{)}^{-2}-2x^2(1-x{)}^{-3}=x+4x^2+9x^3+\cdots +n^2{x}^n+\dots$ - and that’s exactly what we’re trying to get to, with $x=\frac{1}{2}$.

The left hand side simplifies to $\frac{x(x+1)}{(1-x{)}^3}$. When $x=\frac{1}{2}$, the top is $\frac{3}{4}$ and the bottom is $\frac{1}{8}$, giving the answer of 6.

I’d be interested to hear of any more intuitive approaches, though!

Hope that helps!

- Uncle Colin