<s>[INST] Proof of the equality $\sum\limits_{k=1}^{\infty} \frac{k^2}{2^k} = 6$ Show that for $k$ running over positive integers 
$$ \sum_{k=1}^\infty \frac{k^2}{2^k}=6 .$$ We can use finite calculus.
 [/INST] Solution
For convenience, we denote $$a_k=\frac{k^2+2k+3}{2^{k-1}},~~~k=1,2,\cdots$$
Then $$\frac{k^2}{2^k}=a_k-a_{k+1}.$$
Thus
\begin{align*}\sum_{k=1}^n \frac{k^2}{2^k}&=(a_1-a_2)+(a_2-a_3)+\cdots+(a_n-a_{n+1})=a_1-a_{n+1}\\&=6-\frac{(n+1)^2+2(n+1)+3}{2^n}\\&=6-\frac{n^2+4n+6}{2^n}.\end{align*}
It follows that \begin{align*}\sum_{k=1}^\infty\frac{k^2}{2^k}&=\lim_{n \to \infty}\sum_{k=1}^n \frac{k^2}{2^k}\\&=\lim_{n \to \infty}\left(6-\frac{n^2+4n+6}{2^n}\right)\\&=6-0\\&=6.\end{align*}
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