Different types of Convergence for a Series Function

Question

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.

Answer 1

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 $f\in L^p([-1,1])$