<s>[INST] Finding $\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$ Finding $$\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$$
I think if $$\int_{1}^{\infty}x^2 \frac{2^{x-1}}{3^x}dx$$
 exists that this sequence is convergent, but I doubt that this integral is equal to the number to which the sequence converges to. I do not know a way to solve this sequence, thought about doing geometric progression, but can't because of $k^2$ 
 [/INST] This is the same as
$$ \frac{1}{2} \sum_{k \geq 1} k^2 \left( \frac{2}{3} \right)^k,$$
which looks almost like the geometric series
$$ f(x) = \frac{1}{2} \sum_{k \geq 1} x^k = \frac{1}{2} \frac{x}{1 - x}$$
at $x = \frac{2}{3}$. Notice that
$$ \frac{d}{dx} f(x) = \frac{1}{2} \sum_{k \geq 1} k x^{k - 1},$$
which is closer to what we want. We want another factor of $k$, so we multiply by $x$ and differentiate again,
$$ \frac{d}{dx} x \frac{d}{dx} f(x) = \frac{1}{2} \sum_{k \geq 1} k^2 x^{k-1}.$$
We are only missing a single factor of $x$, so we multiply by $x$ again to see that
$$ \left( x \frac{d}{dx} \right)^2 f(x) = \frac{1}{2} \sum_{k \geq 1} k^2 x^k.$$
To evaluate, you need only to compute
$$ \left( x \frac{d}{dx} \right)^2 \frac{1}{2} \frac{x}{1 - x}$$
and evaluate it at $x = \frac{2}{3}$.
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