INFITE SUM 1/(4n+1)(4n+2)(4n+3)(4n+4) AND (-1)^n/(4n+1)(4n+2)(4n+3)(4n+4)

Following the previous two posts 1,2, today we will compute the following two infinite sums, probably the last post about these sort of sums.

$\sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{\ln 2}{4} - \frac{\pi}{24}$

$S=\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{\pi}{6\sqrt{2}}-\frac{\ln 2}{24}-\frac{\pi}{16}-\frac{\coth^{-1} \sqrt{2}}{6\sqrt{2}}$

As before, the first sum is pretty straightforward, but the second one is much harder and demanded a good amount of computations. The first sum I could find the close form result to check our answer. The second one I could not find anywhere, but checked Wolfram Infinite sum calculator and it matches numerically!

$S=\sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}$

$=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+4)}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(4n+2)(4n+3)}$

$=\frac{1}{2}\cdot\frac{1}{3}\sum_{n=0}^{\infty}\left( \frac{1}{(4n+1)}-\frac{1}{(4n+4)}\right)-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{(4n+2)}-\frac{1}{(4n+3)}\right)$

$=\frac{1}{6}\cdot\frac{1}{4}\sum_{n=0}^{\infty}\left( \frac{1}{n+\frac{1}{4}}-\frac{1}{n+1}\right)-\frac{1}{2}\cdot\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{1}{n+\frac{1}{2}}-\frac{1}{n+\frac{3}{4}}\right)$

$=-\frac{1}{24}\left( \psi\left(\frac{1}{4}\right)+\gamma\right)-\frac{1}{8}\sum_{n=0}^{\infty}\left(\psi\left(\frac{3}{4}\right)-\psi\left(\frac{1}{2}\right)\right)$

The values of $\psi\left(\frac{1}{2}\right)\,\,\text{,}\psi\left(\frac{1}{4}\right)\,\,\text{and}\,\,\psi\left(\frac{3}{4}\right)$ can be found here, then

$S=-\frac{1}{24}\left(-\frac{\pi}{2}-3 \ln 2-\gamma+\gamma\right)-\frac{1}{8}\left(\frac{\pi}{2}-3 \ln 2-\gamma+\gamma+2 \ln 2\right)$

$=\frac{\pi}{38}+\frac{\ln 2}{8}-\frac{\pi}{16}+\frac{3 \ln 2}{8}-\frac{\ln 2}{4}$

$\boxed{\sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{\ln 2}{4} - \frac{\pi}{24}}$

$S=\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)(4n+2)(4n+3)(4n+4)}$

$=\frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)(4n+4)}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(4n+2)(4n+3)}$

$=\frac{1}{2}\cdot\frac{1}{3}\sum_{n=0}^{\infty}\left( \frac{(-1)^n}{4n+1}-\frac{(-1)^n}{4n+4}\right)-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{4n+2}-\frac{(-1)^n}{4n+3}\right)$

$=\frac{1}{6}\sum_{n=0}^{\infty}(-1)^n\left(\int_0^1x^{4n}dx-\int_0^1x^{4n+3}dx \right)-\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\left(\int_0^1x^{4n+1}dx-\int_0^1x^{4n+2}dx\right)$

$=\frac{1}{6}\left(\int_0^1\sum_{n=0}^{\infty}(-x^4)^ndx-\int_0^1x^{3}\sum_{n=0}^{\infty}(-x^4)^ndx \right)-\frac{1}{2}\left(\int_0^1x\sum_{n=0}^{\infty}(-x^4)^ndx-\int_0^1x^{2}\sum_{n=0}^{\infty}(-x^4)^ndx\right)$

$=\frac{1}{6}\left(\int_0^1\frac{1}{1+x^4}dx-\int_0^1\frac{x^{3}}{1+x^4}dx \right)-\frac{1}{2}\left(\int_0^1\frac{x}{1+x^4}dx-\int_0^1\frac{x^{2}}{1+x^4}dx\right)$

Plugging the results (A.6),(A.9),(A.8) and (A.4) in the above equation we obtain

$S=\frac{1}{6}\left(\frac{\pi}{4\sqrt{2}}+\frac{1}{2\sqrt{2}}\coth^{-1} \sqrt{2}-\frac{\ln 2}{4}\right)-\frac{1}{2}\left(\frac{\pi}{8}-\frac{\pi}{4\sqrt{2}}+\frac{1}{2\sqrt{2}}\coth^{-1} \sqrt{2}\right)$

And finally!

$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{\pi}{6\sqrt{2}}-\frac{\ln 2}{24}-\frac{\pi}{16}-\frac{\coth^{-1} \sqrt{2}}{6\sqrt{2}}}$

Appendix

Lets start computing the following indefinite integral

$I=\int\frac{1+x^2}{1+x^4}dx=\int\frac{1+x^2}{1+x^4}\cdot \frac{x^{-2}}{x^{-2}}dx$

$=\int\frac{1+x^{-2}}{x^{-2}+x^2}dx$

$=\int\frac{1+x^{-2}}{\left(x-x^{-1}\right)^2+2}dx$

Let $t=x-x^{-1} \,\, \Rightarrow \,\,dt=\left(1+x^{-2}\right)dx$ , then

$I=\int\frac{1}{2+t^2}dt$

$=\frac{1}{2}\int\frac{1}{1+\frac{t^2}{2}}dt$

Let $t=\sqrt{2}u \,\, \Rightarrow \,\,dt=\sqrt{2}du$ , then

$I=\frac{\sqrt{2}}{2}\int\frac{1}{1+u^2}du$

$I=\frac{\sqrt{2}}{2} \arctan u+C$

$\boxed{\int\frac{1+x^2}{1+x^4}dx=\frac{\sqrt{2}}{2} \arctan \left(\frac{x-x^{-1}}{\sqrt{2}} \right)+C}$ (A.1)

$\int\frac{x^2-1}{1+x^4}dx=\int\frac{x^2-1}{1+x^4}\cdot \frac{x^{-2}}{x^{-2}}dx$

$=\int\frac{1-x^{-2}}{x^{-2}+x^2}dx$

$=\int\frac{1-x^{-2}}{\left(x^1+x^{-1}\right)^2-2}dx$

Let $t=x+x^{-1} \,\, \Rightarrow \,\,dt=\left(1-x^{-2}\right)dx$ , then

$I=\int\frac{1}{t^2-2}dt$

$I=\frac{1}{2}\int\frac{1}{\frac{t^2}{2}-1}dt$

Let $t=\sqrt{2}u \,\, \Rightarrow \,\,dt=\sqrt{2}du$ , then

$I=\frac{\sqrt{2}}{2}\int\frac{1}{u^2-1}dt=\frac{\sqrt{2}}{4}\left(\int\frac{1}{u-1}dt-\int\frac{1}{u+1}dt \right)$

$I=-\frac{\sqrt{2}}{4}\ln \frac{u+1}{u-1}+C$

$I=-\frac{\sqrt{2}}{4}\ln \left( \frac{\frac{x+x^{-1}}{\sqrt{2}}+1}{\frac{x+x^{-1}}{\sqrt{2}}-1}\right)+C$

$\boxed{\int\frac{x^2-1}{1+x^4}dx=-\frac{\sqrt{2}}{4}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right)+C}$ (A.2)

$\int\frac{x^2}{1+x^4}dx=\frac{1}{2}\int\frac{2x^2}{1+x^4}dx$

$=\frac{1}{2}\int\frac{x^2+x^2+1-1}{1+x^4}dx$

$=\frac{1}{2}\int\frac{1+x^2}{1+x^4}dx+\frac{1}{2}\int\frac{x^2-1}{1+x^4}dx$ (A.3)

Plugging (A.1) and (A.2) in (A.3)

$\boxed{\int\frac{x^2}{1+x^4}dx=\frac{1}{2\sqrt{2}} \arctan \left(\frac{x^2-1}{\sqrt{2}x} \right)-\frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right)+C}$

$\int_0^1\frac{x^2}{1+x^4}dx=\left[\frac{1}{2\sqrt{2}} \arctan \left(\frac{x^2-1}{\sqrt{2}x} \right)-\frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right)\right]_0^1$

$=\frac{\pi}{4\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln \left( \frac{2+\sqrt{2}}{2-\sqrt{2}}\right)$

$=\frac{\pi}{4\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln \left( \frac{2+\sqrt{2}}{2-\sqrt{2}}\cdot\frac{2^{-1}}{2^{-1}}\right)$

$=\frac{\pi}{4\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln \left( \frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}}\right)$

Applying formula (A.11) we obtain

$\boxed{\int_0^1\frac{x^2}{1+x^4}dx=\frac{\pi}{4\sqrt{2}}-\frac{1}{2\sqrt{2}}\coth^{-1} \sqrt{2}}$ (A.4)

$I=\int\frac{1}{1+x^4}dx$

Note that $2 = (1 + x^{2}) + (1-x^{2})$

$\int\frac{1}{1+x^4}dx=\frac{1}{2}\int\frac{( x^{2}+1) - (x^{2}-1)}{1+x^4}dx$

$=\frac{1}{2}\int\frac{1+x^2}{1+x^4}dx-\frac{1}{2}\int\frac{x^2-1}{1+x^4}dx$ (A.5)

Plugging (A.1) and (A.2) in (A.5) we obtain

$\boxed{\int\frac{1}{1+x^4}dx=\frac{1}{2\sqrt{2}} \arctan \left(\frac{x^2-1}{\sqrt{2}x} \right)+\frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right)+C}$

$\int_0^1\frac{1}{1+x^4}dx=\left[\frac{1}{2\sqrt{2}} \arctan \left(\frac{x^2-1}{\sqrt{2}x} \right)+\frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right)\right]_0^1$