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 $k^2$ 

The sequence can be written as $=\frac{1}{2}\sum_{1}^{\infty} k^2.a^k$ where $a=\frac{2}{3}$
Let $$S = \sum_{1}^{\infty} k^2.a^k\tag{1}$$
Multiply (1) by a,
$$aS = a^2+4a^3+9a^4+\cdot+\infty$$
Subtract now 
$$(1-a)S = a+3a^2+5a^3+\cdot+\infty = 2a-a + 4a^2-a^2 + 6a^3-a^3 +\cdot+\infty$$
$$(1-a)S = 2a(1+2a+3a^2+\cdot+\infty) - a(1+a+a^2+\cdot+\infty)$$
$$(1-a)S = 2aS'-\frac{a}{1-a}$$
$$S' = 1+2a+3a^2+4a^3+\cdot+\infty$$
$$aS' = a+2a^2+3a^3+\cdot+\infty$$
$$(1-a)S' = 1+a+a^2+a^3+\cdot+\infty$$
$$S' = \frac{1}{(1-a)^2}$$
$$(1-a)S = \frac{2a}{(1-a)^2} - \frac{a}{1-a}$$
$$S = \frac{2a}{(1-a)^3} - \frac{a}{(1-a)^2}$$
Substituting the value of $a = \frac{2}{3}$
The sequence equals $\frac{1}{2}\sum_{1}^{\infty} k^2.a^k= \boxed{15}$
