Dear Uncle Colin,

Apparently, $\sum_1^\infty \left(\frac{n^2}{2^n}\right) = 6$, 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 $1 + 2x + 3x^2 + \dots + (n+1)x^n + \dots$.

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

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

Multiply by $x$ again: $x(1-x)^{-2} - 2x^2(1-x)^{-3} = x + 4x^2 + 9x^3 + \dots + n^2 x^n + \dots$ - 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