1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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! $\endgroup$
    – imranfat
    May 6 '13 at 19:18
  • $\begingroup$ 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? $\endgroup$ May 6 '13 at 19:50
2
$\begingroup$

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 $x\in(-1,1)$.

$\endgroup$
9
  • $\begingroup$ Well, now we have the answer....wasn't a tip in the right direction a better move? $\endgroup$
    – imranfat
    May 6 '13 at 19:19
  • 1
    $\begingroup$ 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. $\endgroup$
    – Clement C.
    May 6 '13 at 19:21
  • $\begingroup$ 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. $\endgroup$ May 6 '13 at 20:04
  • 2
    $\begingroup$ I don't understand what you mean by "uncomfortable": this is a closed-form expression -- what did you expect? $\endgroup$
    – Clement C.
    May 6 '13 at 20:35
  • 1
    $\begingroup$ Yes, but the argument of $f$ is the $x$ in the sum, not $n$. $f(0)=\sum_{n=1}^\infty \frac{0^{4n +1}}{4n +1}$... $\endgroup$
    – Clement C.
    May 6 '13 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.