I really have no idea how to start this one!Find with proof an interval on which $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}(1+ \sin x)^n$ determines a differentiable function of x and show that the derivative is $\displaystyle -\cot x$.

My first attempt involved finding when it converges.

$\displaystyle \left|\frac{a_{n+1}}{a_n} \right|= \left| \frac{1}{n+1}(1+ \sin x)^{n+1}. \frac{n}{(1+ \sin x)^n} \right|=\left| \frac{n}{n+1}(1+ \sin x) \right|$

So it converges when $\displaystyle \left| \frac{n}{n+1}(1+ \sin x) \right|<1$.

But $\displaystyle \left| \frac{n}{n+1}(1+ \sin x) \right| \rightarrow |1+ \sin x|$

$\displaystyle |1+ \sin x| \leq 1+ | \sin x|<1 \Rightarrow | \sin x |<0$ which is when $\displaystyle x=n \pi$.

However, I have no idea how to finish it.

Does anyone have any ideas?