I Want to find the function represented by $\sum_{k=1}^{\infty} k^2z^k$.
Here is what I have done so far:
starting with $\sum_{k=1}^{\infty}z^k = \frac{z}{1-z}$ as a definition of a geometric series. We then can take the derivative of the series and multiply it by z yielding $\sum_{k=1}^{\infty}kz^k = \frac{z}{(1-z)^2}$. Repeating this process we obtain $\sum_{k=1}^{\infty}k^2z^k = \frac{z}{(1-z)^2} + \frac{2z^2}{(1-z)^3}$.
Does the index have to change when we integrate this, or is this the final solution and nothing changes?
There is a way to avoid this question of reindexing.
I shall make it a bit more difficult with $\sum_{k=1}^{\infty} k^3z^k$.
Write $$k^3=k(k-1)(k-2)+a k(k-1)+b k +c$$ that is to say $$c+k (-a+b+2)+(a-3) k^2=0$$ from which $a=3$, $b=1$ and $c=0$. So $$\sum_{k=1}^{\infty} k^3z^k=\sum_{k=1}^{\infty}k(k-1)(k-2)z^k+3\sum_{k=1}^{\infty}k(k-1)z^k+\sum_{k=1}^{\infty}kz^k$$ which can write $$\sum_{k=1}^{\infty} k^3z^k=z^3\sum_{k=1}^{\infty}k(k-1)(k-2)z^{k-3}+3z^2\sum_{k=1}^{\infty}k(k-1)z^{k-2}+z\sum_{k=1}^{\infty}kz^{k-1}$$ that is to say $$\sum_{k=1}^{\infty} k^3z^k=z^3\left(\sum_{k=1}^{\infty}z^k\right)'''+3z^2\left(\sum_{k=1}^{\infty}z^k\right)''+z\left(\sum_{k=1}^{\infty}z^k\right)'$$