Find which differentiable function is determined by this power series

Question

Given the series $$\sum\limits_{n=1}^\infty \dfrac{1}{n(n+1)} (\sin(x))^n$$ for all $x \in \mathbb{R}$, find an interval on which it determines a differentiable function of $x$, together with an expression for its derivative in terms of standard function.

My Attempt

Using comparison test , we can see that the above series converges for all $|\sin(x)| \leq 1 \iff \forall x \in \mathbb{R}$.
now if we differentiate the above series we get; $$\sum\limits_{n=1}^\infty \dfrac{\cos(x)}{(n+1)} (\sin(x))^{n-1}$$ which if we rewrite it is;
$$\cos(x) \sum\limits_{n=0}^\infty \dfrac{1}{(n+2)} (\sin(x))^{n}$$

But I can't make out what function this power series relates to?

Any help would be much appreciated.

Answer 1

Hint

Rewrite, using $y=\sin(x)$ and partial fraction decomposition, $$\sum\limits_{n=1}^\infty \dfrac {\sin(x)^n}{n(n+1)}=\sum\limits_{n=1}^\infty \dfrac {y^n}{n(n+1)}=\sum\limits_{n=1}^\infty \dfrac {y^n}{n}-\dfrac {1}{y}\sum\limits_{n=1}^\infty \dfrac {y^{n+1}}{n+1}$$ Now recognize that each of the sum is the antiderivative of quite well known sums which are simple geometric progressions.

When you finish the work with these two sums, replace $y$ by $\sin(x)$. You will end with a rather simple expression.

I am sure that you can take from here

Answer 2

It turns into a differential equation. Can you solve the ODE you get by your method?

EDIT:

Consider $f'(x)+\cos(x)f(x)$:

$$\frac{\cos(x)}{2}+\cos(x)\sum_{n=1}^{\infty}\frac{1}{n+2}(\sin(x))^n+\cos(x)\sum_{n=1}^{\infty}\frac{1}{n(n+1)}(\sin(x))^n$$

$$=\frac{\cos(x)}{2}+\cos(x)\sum_{n=1}^{\infty}(\frac{1}{n+2}-\frac{1}{n(n+1)})(\sin(x))^n$$

Notice that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

So, you can get rid of all powers of $n$ that are greater than $1$ in the denominator.

Then use the identity:

$$\frac{1}{1-t}=1+t+t^2+\cdots \text{ for } |t|<1$$

to derive the formulas you need for your power series. Notice that if we set $t=\sin(x)$ things become clearer.

Can you take it from here?

Second EDIT:

Actually I just realized that turning it into an ODE doesn't simplify things. Even though it works, but it's just a step that is unnecessary.

Just use partial fractions the same way as explained.

Answer 3

Consider: \begin{align} \sum_{n \ge 0} u^n &= \frac{1}{1- u} \\ \int_0^u \sum_{n \ge 0} t^n \, \mathrm{d} t &= \sum_{n \ge 1} \frac{t^n}{n} \\ &= -\ln (1 - u) \end{align} Another round of integration gets the right denominator, divide by $u$ for the numerator, and all this is valid for $\lvert u \rvert < 1$ (with a removable singularity at 0). Substitute $\sin x$ when you are done.