I am trying to evaluate the following integral:
$\lim_{n \to \infty} \int_{(0,1)} \sum^{2n+1}_{k=0} (-1)^{k} x^{2k} dm(x)$
Attempt:
Let $f_{n}(x)=\sum^{2n+1}_{k=0} (-1)^{k} x^{2k}$
Then $f_{n}(x)$ can be written as:
$f_{n}(x)=1-x^{2}+x^{4}-...+x^{4n}-x^{4n+2}$ $=(1-x^{2})(1+x^{4}+...+x^{4n})$
Now for all $x \in (0,1)$, $(1-x^{2})>0$ and $(1+x^{4}+...+x^{4n})>1$
So $f_{n}(x)$ is a sequence of non negative functions that are increasing as $f_{n+1}(x) =f_{n}(x)+(1-x^{2})x^{4n+4}>f_{n}(x)$
Then we can use the monotone convergence theorem:
$\lim_{n \to \infty} \int_{(0,1)} \sum^{2n+1}_{k=0} (-1)^{k} x^{2k} dm(x)$ $ = \int_{(0,1)} \lim_{n \to \infty} \sum^{2n+1}_{k=0} (-1)^{k} x^{2k} dm(x)$ $ = \int_{(0,1)} \sum^{\infty}_{k=0} (-1)^{k} x^{2k} dm(x)$
The function is bounded and continuous over (0,1), so the Lebesgue and Riemann integrals coincide. In which case we have:
$ \int_{(0,1)} \sum^{\infty}_{k=0} (-1)^{k} x^{2k} dm(x) = \int_{0}^{1} \sum^{\infty}_{k=0} (-1)^{k} x^{2k} dx$
Now I'm not sure if what I've done up to this point is correct, and I'm not sure how to proceed from here. Integrating the above expression gets me $\sum^{\infty}_{k=0} \frac{(-1)^{k}}{2k+1}$ which I'm not sure how to evaluate. But the following question says to use the result from this question to evaluate the sum $\sum^{\infty}_{k=0} \frac{(-1)^{k}}{2k+1}$ which makes me think that there is an alternative way to calculate the integral that will allow me to evaluate the sum?
Any help would be appreciated!
For $x\lt 1$, $\sum_{k=0}^\infty(-1)^kx^{2k}=\frac{1}{1+x^2}$. Then $\int_0^1 \frac{1}{1+x^2}dm(x)=\frac{\pi}{4}$ To be rigorous, you need to show $\int_{1-u}^1\sum_{k=0}^\infty(-1)^kx^{2k}dm(x)\to 0$, as $u\to 0$.