INFINITE SUM FROM GRADSHTEYN AND RYZHIK I

I saw the following infinite sum in the monumental Table of Integrals, Series, and Products of Gradshteyn and Ryzhik in entry 0.238 3. and wanted to proof it.

$\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)(3n+4)}=\frac{1}{6}-\frac{ \ln 3}{4}+\frac{\pi}{12\sqrt{3}}$

Since it wasn´t that bad, I thought of a variation, namely

$\sum_{n=0}^{\infty}\frac{(-1)^n}{(3n+1)(3n+2)(3n+3)(3n+4)}= -\frac{1}{6}+\frac{4 \ln 2}{9}-\frac{\pi \sqrt{3}}{54}$

This one was way harder than the first, but unfortunately it does not appear in GR to check if my computation matched with the correct answer. Then, I went to the infinite sum calculator of Wolfram, and was lucky enough that for this sum it returned back a closed form that matched with the calculations!

So lets get to it!

We will rely on the Digamma function to evaluate both sums, so it´s good to recall the following properties:

$\psi(x)=-\gamma+\sum_{k=0}^{\infty}\left(\frac{1}{(n+1)}-\frac{1}{\left(n+x\right)} \right)$ (1)

$\psi(x)-\psi(y)=\sum_{k=0}^{\infty}\left(\frac{1}{(n+y)}-\frac{1}{\left(n+x\right)} \right)$ (2)

$\psi(1+x)=\frac{1}{x}+\psi(x)$ (3)

For the first sum the biggest challenge is the partial fraction decomposition. Once We get through it, the Digamma function technique take´s care.

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

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

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

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

$=\frac{1}{18} \sum_{n=0}^{\infty}\left(\frac{1}{\left(n+\frac{1}{3}\right)}-\frac{1}{\left(n+\frac{4}{3}\right)} \right)+ \frac{1}{6}\sum_{n=0}^{\infty}\left(\frac{1}{(n+1)}-\frac{1}{\left(n+\frac{2}{3}\right)} \right)$

$=\frac{1}{18} \left(\psi\left(\frac{4}{3}\right)-\psi\left(\frac{1}{3}\right) \right)+ \frac{1}{6}\left(\gamma+\psi\left(\frac{2}{3}\right)\right)$

From equation (1),(2) and (3) above and this post we get

$S=\frac{1}{18} \left(3-\frac{\pi}{2\sqrt{3}}-\frac{3 \ln 3}{2}-\gamma+\frac{\pi}{2\sqrt{3}}+\frac{3 \ln 3}{2}+\gamma\right)+ \frac{1}{6}\left(\gamma+\frac{\pi}{2\sqrt{3}}-\frac{3 \ln 3}{2}-\gamma\right)$

And finally!

$\boxed{\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)(3n+4)}=\frac{1}{6}-\frac{ \ln 3}{4}+\frac{\pi}{12\sqrt{3}}}$ (4)

The second sum requires a little more computation and patience but it´s worth it for sure. The partial fraction decomposition, is exactly the same as above.

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

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

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

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

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

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

$= \frac{1}{6}\int_0^1\frac{1-x^{3}}{1+x^3}dx-\frac{1}{2}\int_0^1\frac{x-x^{2}} {1+x^3}dx$

$= \frac{1}{6}\left(\int_0^1\frac{1}{1+x^3}dx-\int_0^1\frac{x^{3}}{1+x^3}dx\right)+\frac{1}{2}\left(\int_0^1\frac{x^{2}} {1+x^3}dx-\int_0^1\frac{x} {1+x^3}dx\right)$ (5)

All integrals in (5) are special instances of the following result proved in the appendix:

$\boxed{\int_0^1\frac{x^{m}}{1+x^n}dx=\frac{1}{2n}\left(\psi\left( \frac{m+n+1}{2n}\right)-\psi\left(\frac{m+1}{2n} \right)\right)}$

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

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

$\boxed{\int_0^1\frac{1}{1+x^3}dx=\frac{\ln 2}{3}+ \frac{\pi \sqrt{3}}{9}}$ (6)

$\int_0^1\frac{x^{3}}{1+x^3}dx=\frac{1}{6}\left(\psi\left( \frac{7}{6}\right)-\psi\left(\frac{2}{3} \right)\right)$

$=\frac{1}{6}\left(6+\psi\left( \frac{1}{6}\right)- \frac{\pi \sqrt{3}}{6}+\frac{3 \ln 3}{2}}+\gamma \right)\right)$

$=\frac{1}{6}\left(6-\frac{\pi \sqrt{3}}{2}-2 \ln 2-\frac{3 \ln 3}{2}-\gamma- \frac{\pi \sqrt{3}}{6}+\frac{3 \ln 3}{2}}+\gamma \right)\right)$

$\boxed{\int_0^1\frac{x^{3}}{1+x^3}dx=1-\frac{\ln 2}{3}- \frac{\pi \sqrt{3}}{9}}$
(7)

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

$=\frac{1}{6}\left(-\gamma+\gamma+2 \ln 2\right)$

$\boxed{\int_0^1\frac{x^{2}}{1+x^3}dx=\frac{\ln 2}{3}}$ (8)

$\int_0^1\frac{x}{1+x^3}dx=\frac{1}{6}\left(\psi\left( \frac{5}{6}\right)-\psi\left(\frac{1}{3} \right)\right)$

$=\frac{1}{6}\left(\frac{\pi\sqrt{3}}{2}-2 \ln 2-\frac{3 \ln 3}{2}-\gamma +\frac{\pi\sqrt{3}}{6}+\frac{3 \ln 3}{2}+ \gamma\right)$

$\boxed{\int_0^1\frac{x}{1+x^3}dx=\frac{\pi \sqrt{3}}{9}-\frac{\ln2}{3}}$ (9)

Plugging (6),(7),(8) and (9) in (5)

$\sum_{n=0}^{\infty}\frac{(-1)^n}{(3n+1)(3n+2)(3n+3)(3n+4)}= -\frac{1}{6}+\frac{\ln 2}{9}+\frac{\pi \sqrt{3}}{27}+\frac{\ln 2}{3}-\frac{\pi \sqrt{3}}{18}$

And finally we obtain the wonderful result:

$\boxed{\sum_{n=0}^{\infty}\frac{(-1)^n}{(3n+1)(3n+2)(3n+3)(3n+4)}= -\frac{1}{6}+\frac{4 \ln 2}{9}-\frac{\pi \sqrt{3}}{54}}$ (10)

Appendix

Recall the integral representation of the Digamma function

$\psi(z)=\int_{0}^{\infty} \frac{e^{-t}}{t}-\frac{e^{-z t}}{1-e^{-t}} d t$

$\text {by the following substitution } e^{-t}=x \Rightarrow d t=-\frac{d x}{x} \,\text {we get }$

$\psi(z)=\int_{1}^{0} \frac{x}{-\ln x}-\frac{x^{z}}{1-x} \frac{(-d x)}{x}$

$\boxed{\psi(z)=-\int_{0}^{1} \frac{1}{\ln x}+\frac{x^{z-1}}{1-x} d x}$ (A.1)

Now lets proof the following result:

$\int_0^1\frac{x^{p-1}-x^{q-1}}{1-x}dx=\psi\left( q\right)-\psi\left(p \right)$

Proof:

$\int_0^1\frac{x^{p-1}-x^{q-1}}{1-x}dx=\int_0^1\frac{x^{p-1}-x^{q-1}}{1-x}+\frac{1}{\ln x}-\frac{1}{\ln x}dx$

$=-\int_{0}^{1} \frac{1}{\ln x}+\frac{x^{q-1}}{1-x} d x+\int_{0}^{1} \frac{1}{\ln x}+\frac{x^{p-1}}{1-x} d x$

$\boxed{\int_0^1\frac{x^{p-1}-x^{q-1}}{1-x}dx=\psi\left( q\right)-\psi\left(p \right)}$ (A.2)

$\int_0^1\frac{x^{m}}{1+x^n}dx=\frac{1}{2n}\left(\psi\left( \frac{m+n+1}{2n}\right)-\psi\left(\frac{m+1}{2n} \right)\right)$

Proof:

$\int_0^1\frac{x^{m}}{1+x^n}dx=\int_0^1\frac{x^{m}}{1+x^n}\frac{(1-x^n)}{(1-x^n)}dx$

$=\int_0^1\frac{x^{m}-x^{m+n}}{1-x^{2n}}dx$

$\text{let}\,\, x^{2n}=u \Rightarrow \,x=u^{\frac{1}{2n}}\Rightarrow \,dx=\frac{1}{2n}u^{\frac{1}{2n}-1}du$

Thus

$\int_0^1\frac{x^{m}}{1+x^n}dx=\frac{1}{2n}\int_0^1\frac{u^{\frac{m+1}{2n}-1}-u^{\frac{m+n+1}{2n}-1}}{1-u}du$

and from (A.2) we can conclude that

$\boxed{\int_0^1\frac{x^{m}}{1+x^n}dx=\frac{1}{2n}\left(\psi\left( \frac{m+n+1}{2n}\right)-\psi\left(\frac{m+1}{2n} \right)\right)}$ (A.3)

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INFINITE SUM FROM GRADSHTEYN AND RYZHIK I

Appendix

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