Calculate $\sum_{n=1} ^{\infty} n (\frac{1+i}{2 \sqrt{2}})^n$ How to calculate $\sum_{n=1} ^{\infty} n (\frac{1+i}{2 \sqrt{2}})^n$ ?
By the ratio test, the series converges, but I don't know how to calculate its limit.
Could you help me?
 A: Recall that $$f(x) = \sum_{n=0}^{\infty} x^n = \dfrac1{1-x}$$
for all $\vert x \vert \leq 1$ and $x \neq 1$. Hence,
$$f'(x) = \sum_{n=0}^{\infty} nx^{n-1} = \dfrac1{(1-x)^2}$$
This gives us
$$\sum_{n=0}^{\infty} nx^{n} = \dfrac{x}{(1-x)^2}$$
Now plug in $x = \dfrac{1+i}{2\sqrt2}$ to get the answer.
A: $$\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}, |z|<1$$
Differentiating both sides w.r.t. $z$ gives,
$$\sum_{n=1}^{\infty}nz^{n-1}=\frac{1}{(1-z)^2}$$ (because this series converges uniformly), now multiply both sides by $z$ gives, 
$$\sum_{n=1}^{\infty}nz^{n}=\frac{z}{(1-z)^2}$$
EDIT: Proof that $\sum_{n=0}^{\infty}z^n$ converges uniformly:
Consider partial sum $s_n=\sum_{k=0}^{n}z^k$ and $s_m=\sum_{k=0}^{m}z^k$
then,for $|m-n|<\delta$ $$|s_n-s_m|=\big|\frac{x^{m+1}-x^{n+1}}{1-x}\big|\leq |x^{m+1}|\big|\frac{1-x^{n-m}}{1-x}\big|\leq \big|\frac{1-x^{\delta}}{1-x}\big|=\epsilon$$ 
Thus, for a given $\epsilon >0$, choose $\delta$ such that $\big|\frac{1-x^{\delta}}{1-x}\big|=\epsilon$ and since that $\delta$ depends only on $\epsilon$, therefore, this series converges uniformly.
