# Different types of Convergence for a Series Function

I am currently investigating the convergence of the following function,

$$f(x)=\sum\limits_{k=1}^{\infty} \dfrac{x^{k}+\sin(k)}{k^{2}}$$

for different "senses". I have shown that $$f(x)$$ converges uniformly and pointwisely for its domain. All I have left to do is investigate whether it converges in the $$L^{p}$$ sense. The problem is, I have absolutely no idea what this means! My knowledge on this $$L^{p}$$ is very minimal and I can not seem to find anything on the www or Rudin. If someone could explain to me what this $$L^{p}$$ business is it would be greatly appreciated. A few sub-questions I have include:

• How does it differ from uniform convergence?
• Any geometrical interpretations?
• I know that $$p\geq1%$$. How does every each value of $$p$$ affect convergence?
• Are there any relationships between each $$L^{p}$$?

If anyone knows of some nice references to my specific questions, that also would be greatly appreciated.

• do you know what the $L^p$ norm is $\|f\|_p=(\int|f|^pdx)^{\frac{1}{p}}$ Commented Apr 9, 2014 at 15:25
• Ok, thanks @ellya. That is a start to my understanding. Anymore help? Commented Apr 9, 2014 at 21:52
• what does f(x) converge to? Commented Apr 9, 2014 at 21:55
• @ellya. It converges, but I do not to what to. The first term, $\frac{x^{k}}{k{2}}$ converges via the ratio test and letting |x| < 1, and the second term converges via the W-M Test. Are you asking for the function that it converges to? I do not know that. Can you help me find it, perhaps? I have proved Uniform and PW convergence via definition of delta-epsilon Commented Apr 9, 2014 at 22:03

Well we can see that $|f(x)|=|\sum_{k=1}^\infty\frac{x^k+sin(k)}{k^2}|\leq\sum_{k=1}^\infty|\frac{x^k+sin(k)}{k^2}|\leq \sum_{k=1}^\infty|\frac{x^k}{k^2}|+\sum_{k=1}^\infty\frac{1}{k^2}=\sum_{k=1}^\infty|\frac{x^k}{k^2}|+\frac{\pi^2}{6}$

Now like you said if we restrict to $x\in[0,\frac{1}{2})$ (You'll see why later, then we have $|f(x)|\leq\sum_{k=1}^\infty|\frac{x^k}{k^2}|+\frac{\pi^2}{6}\leq \sum_{k=1}^\infty x^k+\frac{\pi^2}{6}=\frac{1}{1-x}+\frac{\pi^2}{6}$

Now as I said $\|f\|_p=(\int |f|^pdx)^{\frac{1}{p}}$.

Let us look at $L^1$, here $\|f\|_1=\int_0^\frac{1}{2}|f(x)|dx=\int_0^\frac{1}{2}\frac{1}{1-x}+\frac{\pi^2}{6}dx=(-\ln(1-x)+\frac{\pi^2}{6}x)|_0^1=-\ln{\frac{1}{2}}+\frac{\pi^2}{6}=\ln(2)+\frac{\pi^2}{6}\lt\infty$, so $f\in L^1$.

This is all i've thought up so far, but I hope it helps a bit.

edit

I've thought of a much better way to do this.

We as before restrict to $|x|\le 1$, then:

$|f(x)|=|\sum_{k=1}^\infty\frac{x^k+sin(k)}{k^2}|\leq\sum_{k=1}^\infty|\frac{x^k+sin(k)}{k^2}|\leq\sum_{k=1}^\infty\frac{2}{k^2}=\frac{\pi^2}{3}$

Thus $\int_{-1}^{1}|f(x)|^pdx\leq\int_{-1}^{1}(\frac{\pi^2}{3})^pdx=2(\frac{\pi^2}{3})^p\lt\infty$, Thus $f\in L^p([-1,1])$

• Hey @ellya, yes that has made more sense. A few questions. What you did in the first 4 lines, what did you achieve? Did you show that my original function converges to $\sum_{k=1}^\infty|\frac{x^k}{k^2}|+\frac{\pi^2}{6}$? If so, I can not see why. Sure, I know the function converges, but would not be satisfied to conclude that it converges to that. Now, the domain is [-1,1]. And I am curious to know why you only went from 0 -> (1/2).How about L2 and L3 and so on? Thanks ! Commented Apr 9, 2014 at 23:04
• Firstly, what I found was an upper bound for f, which allowed me to show that f converges under the $L^1$ norm. I took $(0,1/2)$ literally because it simplifies this integral, Im sure with more effort you can extend the interval. As for L^p in general, the integral just becomes a bit more complicated, as we take the pth power in the integral. Commented Apr 9, 2014 at 23:11
• I see. With the $L^{p}$ is it possible though? Because I am required to see if it converges in the $L^{p}$ sense. So, how can I come to a final conclusion? Any hint - I don't expect you to do it for me. Commented Apr 9, 2014 at 23:18
• Try what I did, firstly on (0,1/2) BUT do the integral of $|f|^p$ and see what happens,hopefully it will be finite, after that try making the interval larger. Commented Apr 9, 2014 at 23:21
• Also what was the domain of f meant to be? Its not in your post. Commented Apr 9, 2014 at 23:24