Divergent Series?

Note that $\displaystyle S = \sum_{k=1}^\infty \frac {k^2}{2^k} = \sum_{k=\color{#D61F06}{0}}^\infty \frac {k^2}{2^k}$ . Then, we have

$\begin{aligned} S & = \sum_{k=1}^\infty \frac {k^2}{2^k} \\ & = \sum_{k=\color{#3D99F6}{0}}^\infty \frac {({\color{#3D99F6}{k+1}})^2}{2^{\color{#3D99F6}{k+1}}} \\ & = \frac 12 \sum_{k=0}^\infty \frac {k^2+2k+1}{2^k} \\ & = \frac 12 \sum_{k=0}^\infty \frac {k^2}{2^k} + {\color{#3D99F6}\sum_{k=0}^\infty \frac k{2^k}} + \frac 12 \sum_{k=0}^\infty \frac 1{2^k} & \small \color{#3D99F6}{\text{See Note}} \\ & = \frac 12 S + {\color{#3D99F6}2} + \frac 12 \cdot \frac 1{1-\frac 12} \\ \implies \frac 12 S & = 2 + 1 \\ \implies S & = \boxed{6} \end{aligned}$

---

Note:

Using the similar method:

$\begin{aligned} S_1 & = \sum_{k=0}^\infty \frac k{2^k} \\ & = \sum_{k={\color{#D61F06}1}}^\infty \frac k{2^k} \\ & = \sum_{k={\color{#3D99F6}0}}^\infty \frac {\color{#3D99F6}k+1}{2^{\color{#3D99F6}k+1}} \\ & = \frac 12 \sum_{k=0}^\infty \frac k{2^k} + \frac 12 \sum_{k=0}^\infty \frac 1{2^k} \\ & = \frac 12 S_1 + \frac 12 \left(\frac 1{1-\frac 12}\right) \\ \implies \frac 12 S_1 & = 1 \\ \implies S_1 & = 2 \end{aligned}$

An algebraic approach, Let

$S=\displaystyle\sum_{k=1}^{\infty} \dfrac{k^2}{2^k}$

$\color{#FFFFFF}{dang}$

$S= \dfrac{1^2}{2^1}+\dfrac{2^2}{2^2}+\dfrac{3^2}{2^3}+\dfrac{4^2}{2^4}+\dfrac{5^2}{2^5}+\dfrac{6^2}{2^6}+\cdots$

$\color{#FFFFFF}{dang}$

$\begin{aligned} 2S & = 1+\dfrac{2^2}{2^1}+\dfrac{3^2}{2^2}+\dfrac{4^2}{2^3}+\dfrac{5^2}{2^4}+\dfrac{6^2}{2^5}+ \cdots \\ -S & \underline{=-\bigg(\dfrac{1^2}{2^1}+\dfrac{2^2}{2^2}+\dfrac{3^2}{2^3}+\dfrac{4^2}{2^4}+\dfrac{5^2}{2^5}+\cdots\bigg)} \end{aligned}$

$S=1+\dfrac{2^2-1^1}{2^1}+\dfrac{3^2-2^2}{2^2}+\dfrac{4^2-3^2}{2^3}+\dfrac{5^2-4^2}{2^4}+\dfrac{6^2-5^2}{2^5}+\cdots$

$\color{#FFFFFF}{dang}$

$S=1+\dfrac{(2+1)(2-1)}{2^1}+\dfrac{(3+2)(3-2)}{2^2}+\dfrac{(4+3)(4-3)}{2^3}+\dfrac{(5+4)(5-4)}{2^4}+\dfrac{(6+5)(6-5)}{2^5}+\cdots$

$S=1+\color{#D61F06}{\dfrac{3}{2^1}+\dfrac{5}{2^2}+\dfrac{7}{2^3}+\dfrac{9}{2^4}+\dfrac{11}{2^5}+\cdots}$

$\color{#FFFFFF}{dang}$

$\qquad\color{#D61F06}{K=\dfrac{3}{2^1}+\dfrac{5}{2^2}+\dfrac{7}{2^3}+\dfrac{9}{2^4}+\dfrac{11}{2^5}+\cdots}$

$\qquad\color{#D61F06}{2K=3+\dfrac{5}{2^1}+\dfrac{7}{2^2}+\dfrac{9}{2^3}+\dfrac{11}{2^4}+\cdots}$

$\qquad\color{#D61F06}{\underline{-K \;\; -\bigg(\dfrac{3}{2^1}+\dfrac{5}{2^2}+\dfrac{7}{2^3}+\dfrac{9}{2^4}+\cdots\bigg)}}$

$\qquad\color{#D61F06}{K=3+}\color{#3D99F6}{\big(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots\big)}$

$\qquad\color{#D61F06}{K=3+}\,\color{#3D99F6}{2}$

$S=1+5$

$S=\huge\boxed{6}$

very nice solution.