Let c be a fixed number and consider the power series $\displaystyle\sum_{n=1}^ \infty \frac{c^{n-1}}{n} x^{n}$.
a) Determine the convergence radius r for every value of $c \in \mathbb{C}$.
In this task I used the ratio test: $ \mid \frac{a_{n+1}}{a_n} \mid $ this gives me the result $ \mid \frac{n}{n+1}cx \mid$ and this will be $ \mid cx \mid$ as $ n \rightarrow \infty $.
And I know that if $ \mid cx \mid$ $ < 1 $ it will converge. The convergence interval will be $ \frac{-1}{c} < x < \frac{1}{c} $.
To show that this is the convergence interval I would just put the values istead of x and see what happens.
To determine the convergence radius r : $ \frac{\frac{1}{c} -(-\frac{1}{c})}{2} =\frac{1}{c}$.
Is this correct??
b) Let $f$: ]-r,r[ $ \rightarrow \mathbb{C} $ describe the sum function to the power series above.
Show for all real c $\neq $ 0 that $f(x)$ is a strictly increasing function of $x$
Do you have any ideas for this task??
Your solution of (a) is correct, although using the formula of Cauchy-Hadamard,
$$ R = \frac{1}{\limsup_{n\rightarrow \infty} \sqrt[n]{|a_n|}}$$
for the power series $\sum_n a_n x^n$ would perhaps have been faster.
For (b) you write your series as $c^{-1} \cdot \sum_n \frac{(cx)^n}{n}$ and then calculate the derivative explicitely (term by term) (why is this legitimate?). Show that the result is non-negative.