# Math Help - Power series

1. ## Power series

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

My first attempt involved finding when it converges.

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

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

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

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

However, I have no idea how to finish it.

Does anyone have any ideas?

2. Remembering the Taylor expansion...

$\sum_{n=1}^{\infty} \frac{\xi^{n}}{n}= - \ln (1-\xi)$ , $-1 \le \xi < 1$ (1)

... and setting $1 + \sin x= \xi$ we obtain...

$\sum _{n=1}^{\infty} \frac{(1+\sin x)^{n}}{n} = -\ln (-\sin x)$, $-\pi < x < 0$ (2)

From (2) results...

$\frac{d}{dx} -\ln (- \sin x)= - \cot x$ (3)

Kind regards

$\chi$ $\sigma$

3. Ah, well that is interesting!

I have one query though:
$

\sum_{n=1}^{\infty} \frac{\xi^{n}}{n}= - \ln (1-\xi)
$
How did you know that this taylor expansion would give the required result? (or can you solve any question of this type using this taylor expansion?)

4. Originally Posted by Showcase_22
Ah, well that is interesting!

I have one query though:

How did you know that this taylor expansion would give the required result? (or can you solve any question of this type using this taylor expansion?)
It's just because your initial series is in the form $\sum_{n\geq 1} \frac{X^n}{n}$, where $X=1+\sin(x)$. So you can nearly recognize it !
But this series converges iff $|X|<1$, which is not the case if $\sin(x)\geq 0$
So I guess this determines the interval you're asked for.