How do I evaluate $\sum_{i=0}^{\infty}\frac{i^N}{4^i} $? Question:

How do I evaluate $\displaystyle\sum_{i=0}^{\infty}\frac{i^2}{4^i} $? In general, how can one evaluate $\displaystyle\sum_{i=0}^{\infty}\frac{i^N}{4^i} $?

 A: Look at
$$
f(z)=\sum_{i=0}^{\infty}z^i=\frac{1}{1-z}.
$$
Now
$$
z\frac{d}{dz}f(z)=z\sum_{i=0}^{\infty}iz^{i-1}=\sum_{i=0}^{\infty}iz^i,
$$
and in general
$$
\left(z \frac{d}{dz}\right)^{k}f(z)=\sum_{i=0}^{\infty}i^kz^i.
$$
To compute your sums, apply the appropriate differential operator to $(1-z)^{-1}$, then evaluate the result at $z=1/4$.
A: ncmathsadist 2 gives you very good link for the answer to your first question. You can continue along the same way he used in order to compute the value of the sum changing the value of the exponent from 2 to 3 and to any number.  
I have not been able to find a way to make the solution totally general, that is to say to express the value of the sum as a function of exponent N.  
Ignoring if your question is or not related to homework, I should mention that the summation if terms (i^N / M^i), for i going from 0 to infinity, is the Hurwitz-Lerch transcendent function with arguments [1/M , -N , 0].
A: Consider
$$\sum_{k=1}^\infty \frac{k^s}{z^k}$$
To reproduce the sum in your question, you can take $s=N$ and $z=4$.
Rewrite as
$$\sum_{k=1}^\infty \frac{(1/z)^k}{k^{-s}}$$
This is exactly
$$\operatorname{Li}_{-s}(1/z)$$
Where we have the polylogarithm.

The polylogarithm obeys the recurrence
$$\operatorname{Li}_{s+1} (z)=\int_0^z \frac{\operatorname{Li}_s(t)}{t}\mathrm dt$$
As well as having the special value
$$\operatorname{Li}_0(z)=\frac{z}{1-z}$$
This leads to the formula
$$\operatorname{Li}_{-n}(z)=\left(z\frac{\partial}{\partial z}\right)^n \frac{z}{1-z}$$
Which is enough to information to construct $\operatorname{Li}_{-n}(z)$ for whatever $n$ you like and then plug in $z=1/4$ or whatever other special value you want.
