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 $k^2$ 

We know that the geometric series converges,
$$\sum_{k = 0}^{\infty} x^k = \frac{1}{1 - x} \;\; |x| < 1$$
Differentiating both sides,
$$\sum_{k = 0}^{\infty} kx^{k-1} = \frac{1}{(1 - x)^2} \;\; |x| < 1$$
Therefore, multiplying throughout by $x$ we get,
$$\sum_{k = 0}^{\infty} kx^{k} = \frac{x}{(1 - x)^2} \;\; |x| < 1$$
Differentiating again we get,
$$\sum_{k = 0}^{\infty} k^2x^{k-1} = \frac{1 + x}{(1 - x)^3} \;\; |x| < 1$$
Multiplying throughout by $x$ we get,
$$\sum_{k = 0}^{\infty} k^2x^{k} = \frac{x(1 + x)}{(1 - x)^3} \;\; |x| < 1$$
Substitute $x = \frac{2}{3} < 1$ to get,
$$\sum_{k = 0}^{\infty} k^2\left(\frac{2}{3}\right)^{k} = 30$$
Divide throughout by $2$ to get,
$$\sum_{k = 0}^{\infty} k^2\frac{2^{k-1}}{3^k} = 15$$
