If $ m ∈ ]0,1[ $ and $ f(x)=\sum_{k>=1}\frac{m^kx^k}{k}$, $-1/m < x < 1/m,$ then $ f'(1)$ equals to:
a) $m$ b) $0$ c) $m/(1+m)$ d) $1/(1-m)$ e) $m/(1-m)$
I tried solving it and got
$$ f(x)=\sum_{k\geq 1}\frac{m^kx^k}{k},$$ $$ f'(x)=\sum_{k\geq 1}\frac{m^kkx^{k-1}}{k},$$ $$ f'(1)=\sum_{k\geq 1}{m^k1^{k-1}}$$
Now, what should I do? Can someone help me, please?
Note that $m\in (0,1)$ and the geometric series is convergent: $$f'(1)=\sum_{k\geq 1}{m^k1^{k-1}}=\lim_{n\to \infty}\sum_{k=1}^{n}{m^k}=\lim_{n\to \infty}\frac{m(1-m^n)}{1-m}=\frac{m}{1-m}.$$