Show that by expanding 1/(1+x^4) as a geometric series and integrating termwise.
We know that $\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in fty}(-1)^nx^{4n}$ so $\displaystyle \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\
&=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}$
So $\displaystyle \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0 ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\
&=\frac{\sqrt{2}\pi}{4}\end{aligned}$
Now what I think you meant to say was this
$\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^2}=\sum_{n=0}^{\in fty}(-1)^nx^{2n}$
So $\displaystyle \begin{aligned}\int_0^1\frac{dx}{1+x^2}&=\int_0^1\ sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}
\end{aligned}$
But note that $\displaystyle \int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}$
So $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$
Also note that the interchanging of all the summations and integrals was due to the integrability of a power seris for all points on the interior of its interval of convergence.
Hmm, but this just simply is not true?
But for the sum look here http://www.mathhelpforum.com/math-he...415-post2.html and note then that your sum should be equal to $\displaystyle \frac{1}{16}\bigg[\frac{\pi}{2\frac{1}{4}\sin\left(\frac{\pi}{4}\rig ht)}-\frac{1}{2\left(\frac{1}{4}\right)^2}\bigg]=\frac{\sqrt{2}\pi-4}{8}$ which is what your first equation indicates.
What do you think is simply not true??
This portion of your first post is my second part of this problem, where i gotta show that the integral from my first post is equal to sqrt(2)*pi/4 like you have shown.
We know that $\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in fty}(-1)^nx^{4n}$ so $\displaystyle \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\
&=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}$
So $\displaystyle \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0 ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\
&=\frac{\sqrt{2}\pi}{4}\end{aligned}$
is there a way to conclude from here to prove