# Sum $\sum\limits_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}$

The Task is to find the Sum of the given function series, that is defined as: $\sum\limits_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}$

I'm kinda lost, but at least i managed to take a few steps towards the solution.

Due to the fact, the given series is a series of functions, i need to determine if the series converges at all. And it does. I found out the Radius of Convergence, its $\lvert x \rvert < 1$

I think, the Sum must be determined somehow this way:

$\lim_{n \to \infty} \sum\limits_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}$

And here is, where im stuck. How do i calculate the Limit of Series of Functions like this?

p.s. the edits were only for improving language and latex

• Hint, why don't you write out a few terms for n=0,1,2. You will recognize a geometric series where x is to be replaced by x^4. Try it out! May 6 '13 at 19:18
• its $(0+\frac{x^5}{5}+\frac{x^9}{9}+\frac{x^{13}}{13}+\frac{x^{17}}{17}\dots)$ and the first derivative is as shown below but i still cannot get the Connection to the given Problem. maybe there is some Kind of a missing link in my brain. What do i have to calculate in order to get the answer? May 6 '13 at 19:50

By properties of power series, the function $f:x\mapsto \sum_{n=0}^\infty \frac{x^{4n+1}}{4n+1}$ is $\mathcal{C}^\infty$ on $(-1,1)$, and $$f^\prime(x) = \sum_{n=0}^\infty x^{4n}$$ Thus, since this last one is easy to compute, $$f^\prime(x) = \frac{1}{1-x^4}$$ for all $x\in(-1,1)$.
• There is still one step to find the solution, since this is only $f^\prime$; furthermore, the hint "derive term-wise" is giving away basically as much. May 6 '13 at 19:21
• But $\frac{1}{1-x^4}$ is not the Sum of $\sum\limits_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}$ that i was looking for? ... otherwise, please give me a hint, on which topic do i have to read, in order to understand why. May 6 '13 at 20:04
• Yes, but the argument of $f$ is the $x$ in the sum, not $n$. $f(0)=\sum_{n=1}^\infty \frac{0^{4n +1}}{4n +1}$... May 6 '13 at 20:51