For what interval does this power series converge and for what interval does it determine a differentiable function?

Question

For what range of values of $x$ does $\sum_{n=1}^{\infty } \dfrac{1}{n}(1+\sin x)^n$ converge?
Find with proof an interval on which it determines a differentiable function of $x$ and show that the derivative is $\cot(x)$.

First Part
Using Ration test, we get that the power series converges for all $$\left|1+\sin(x)\right|<1$$
which implies
$$-2<\sin(x)<0$$

Second Part
Now here I am not exactly too sure on how to proceed;
Differentiating the given function gives us
$$\sum_{n=1}^{\infty }(1+\sin x)^{n-1} = \sum_{n=0}^{\infty }(1+\sin x)^n$$
But how do I show that this is $\cot(x)$?

Answer 1

Also, $$ \sum_{n=1}^{\infty } \dfrac{1}{n}(1+\sin x)^n=-\ln(1-(1+\sin(x))=-\ln(-\sin(x)) $$ which again is valid for $\sin(x)<0$.