the sum of $1-\frac{1}{5}+\frac{1}{9}-\frac{1}{13}+.....$

Question

I thought this was the real part of the series: $\sum_{n=0}^\infty \frac{i^n}{1+2n}$, with $i=\sqrt{-1}$. When taking the real part I am left with: $\sum_{n=0}^\infty \frac{\cos(n\pi/2)}{1+2n}$. I know this sum is around 0.866973, but I have no idea how to come to this answer. Can someone please help me?

Answer 1

Can you do something with $\frac{1}{1+x^4}=1-x^4+x^8-x^{12}+\cdots$, and then integrate term by term and plug in $x=1$ to get your series? (Along the lines of what is done to get $1-\frac{1}{3}+\frac15-\cdots=\frac{\pi}{4}$)

Answer 2

The answer is $\frac{\sqrt2}{8}$ multiplied by $\ln(2+\sqrt2)-\ln(2-\sqrt2)-2\arctan(1-\sqrt2)+2\arctan(1+\sqrt2)$ It is based on the anti-derivative of $\frac{1}{x^4+1}$ which could be done through partial fraction decomposition. The denom factors $(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)$ It is tedious but doable.

The numerators are of the form $Ax+B$ and $Cx+D$ because the denom's are prime.

Plug in 1 for the anti-derivative to arrive at your answer. Give it a try. (allocate half an hour or so, I had done it a while ago and retrieved the paper)

Answer 3

We can write $$\displaystyle I = 1-\frac{1}{5}+\frac{1}{9}-\frac{1}{13}+......\infty = \int_{0}^{1}\left(1-x^{4}+x^{8}-x^{12}+.......\infty\right)$$

So we get $$\displaystyle I =\int_{0}^{1}\frac{1}{1+x^4}dx=\frac{1}{2}\int_{0}^{1}\frac{(x^2+1)-(x^2-1)}{x^4+1}dx =\frac{1}{2}\int_{0}^{1}\frac{x^2+1}{x^4+1}dx-\frac{1}{2}\int_{0}^{1}\frac{x^2-1}{x^4+1}dx$$

Now Let $$\displaystyle J = \int_{0}^{1}\frac{x^2+1}{x^4+1}dx = \int_{0}^{1}\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{2}\right)^2}dx = \frac{1}{\sqrt{2}}\left[\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)\right]_{0}^{1}$$

So we get $$\displaystyle J = \frac{1}{\sqrt{2}}\left(0+\frac{\pi}{2}\right) = \frac{\pi}{2\sqrt{2}}$$

Similarly Let $$\displaystyle K = \int_{0}^{1}\frac{x^2-1}{x^4+1}dx = \int_{0}^{1}\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-\left(\sqrt{2}\right)^2}dx =\frac{1}{2\sqrt{2}}\left[\ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|\right]_{0}^{1}$$

So we get $$\displaystyle K = \frac{1}{2\sqrt{2}}\left[\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)\right] = -\frac{1}{2\sqrt{2}}\left[\ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)\right]=-\frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)$$

So we get $$\displaystyle I = \frac{1}{2}J-\frac{1}{2}K = \frac{\pi}{4\sqrt{2}}+\frac{1}{2\sqrt{2}}\ln(\sqrt{2}+1) = {\frac{\pi+2\ln(\sqrt{2}+1)}{4\sqrt{2}}}$$