I'm trying to determine the limit of the sum
$\lim_{n\to\infty} \sum\limits_{k=1}^n k^2/2^k$
Doing the convergence test shows the sum converges $\lim_{n\to\infty} \frac{(k+1)^2/2^{k+1}}{k^2/2^k} = 1/2$ so there must be a solution, but the normal tricks used to solve geometric series or telescoping series don't seem to work.
Observe that by just differentiating
$$ 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1, \tag2 $$ with respect to $x$ and by multiplying by $x$, we get the identity $$ \sum_{k=1}^{n}kx^k=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x} , \quad |x|<1,\tag3$$ differentiating once more and multiplying by $x$ gives, as $n \to \infty$, using $|x|<1$: $$ \sum_{k=1}^{\infty}k^2x^k=\frac{x (1+x)}{(1-x)^3} , \quad |x|<1,\tag4$$ then put $x:=\dfrac12$ to obtain $$ \sum_{k=1}^{n}k^2/2^k \longrightarrow 6, \quad \text{as} \, n \to \infty. $$
Let $$ S = \sum_{k\geq 1}\frac{k^2}{2^k}. $$ Then: $$ 2S = \sum_{k\geq 1}\frac{k^2}{2^{k-1}} = 1+\sum_{k\geq 1}\frac{(k+1)^2}{2^k},$$ hence: $$ S = 2S-S = 1+\sum_{k\geq 1}\frac{2k+1}{2^k} = 2+2\sum_{k\geq 1}\frac{k}{2^k}$$ but with the same technique, by letting $T=\sum_{k\geq 1}\frac{k}{2^k}$, we have: $$ T=2T-T = 1+\sum_{k\geq 1}\frac{1}{2^k} = 2$$ from which $S=\color{red}{6}$ follows.
One way to do this is find the function $f(r)$ such that $$\dfrac{r^2}{2^r}=f(r-1)-f(r)$$ which is of the form $\dfrac{1}{2^r}\left(Ar^2+Br+C\right).$ Then easily you can find the partial sum $\sum_{k=1}^n\dfrac{k^2}{2^k}$and take the limit $n\to\infty.$
Another way to do this is use the identity $$1+x+x^2+\cdots=\dfrac{1}{1-x},\,\,\,\,|x|\lt 1.$$ Differentiate this power series multiply once by $x$ and again differentiate. Then you have the required answer.
