# Finding Sum of $\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty$

Finding Sum of $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty\; \bf{terms}$$

Try: Writting it as $$\sum^{\infty}_{r=0}\frac{(4r)!}{(4r+4)!}=\sum^{\infty}_{r=0}\frac{1}{(4r+1)(4r+2)(4r+3)(4r+4)}$$

$$\frac{1}{3}\sum^{\infty}_{r=0}\bigg[\frac{1}{(4r+1)(4r+2)(4r+3)}-\frac{1}{(4r+2)(4r+3)(4r+4)}\bigg]$$

$$\frac{1}{6}\bigg[\bigg(\frac{1}{(4r+1)(4r+2)}-\frac{1}{(4r+2)(4r+3)}\bigg)-\bigg(\frac{1}{(4r+2)(4r+3)}-\frac{1}{(4r+3)(4r+4)}\bigg)\bigg]$$

I did not understand how can i solve further, thanks

• $\frac{(4r)!}{(4r+4)!}=\frac{1}{(4r+1)(4r+2)(4r+3)\color{red}{(4r+4)}}$ – JMoravitz May 11 '18 at 3:23
• I don't think there should be an $\infty$ in the sum. – Lord Shark the Unknown May 11 '18 at 3:26
• The partial fraction decomposition is $\frac1{6(4x+1)}-\frac1{2(4x+2)}+\frac1{2(4x+3)}-\frac1{6(4x+4)}$ – Sonal_sqrt May 11 '18 at 3:27

As Piyush Divyanakar remarks, the sum is $$\frac16\sum_{r=0}^\infty\left( \frac{1}{4r+1}-\frac{3}{4r+2}+\frac{3}{4r+3}-\frac{1}{4r+4}\right).$$ This equals $$\frac16\sum_{r=0}^\infty\int_0^1(t^{4r}-3t^{4r+1}+3t^{4r+2}-t^{4r+3})\,dt =\frac16\int_0^1\frac{(1-t)^3}{1-t^4}\,dt.$$ You just have to do this integral...
Another pssible way is to consider partial sums $$S_p=\frac16\sum_{r=0}^p\left( \frac{1}{4r+1}-\frac{3}{4r+2}+\frac{3}{4r+3}-\frac{1}{4r+4}\right)$$ and to use generalized harmonic numbers $$\sum_{r=0}^p\frac{1}{4r+1}=\frac{1}{8} \left(2 H_{p+\frac{1}{4}}+\pi +\log (64)\right)$$ $$\sum_{r=0}^p\frac{1}{4r+2}=\frac{1}{4} \left(H_{p+\frac{1}{2}}+\log (4)\right)$$ $$\sum_{r=0}^p\frac{1}{4r+3}=\frac{1}{4} \left(H_{p+\frac{3}{4}}-\frac{\pi }{2}+\log (8)\right)$$ $$\sum_{r=0}^p\frac{1}{4r+4}=\frac{H_{p+1}}{4}$$ which make $$S_p=\frac{1}{24} \left(H_{p+\frac{1}{4}}-3 H_{p+\frac{1}{2}}+3 H_{p+\frac{3}{4}}-H_{p+1}-\pi +\log (64)\right)$$ Now, using the asymptotics, $$S_p=\frac{1}{24} (\log (64)-\pi )-\frac{1}{768 p^3}+O\left(\frac{1}{p^4}\right)$$ wcih shows the limit and how it is approached.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{\infty}{\pars{4k}! \over \pars{4k + 4}!} & = {1 \over 6}\sum_{k = 0}^{\infty} {\Gamma\pars{4k + 1}\Gamma\pars{4} \over \Gamma\pars{4k + 5}} = {1 \over 6}\sum_{k = 0}^{\infty} \int_{0}^{1}t^{4k}\pars{1 - t}^{3}\,\dd t \\[5mm] & = {1 \over 6}\int_{0}^{1}\pars{1 - t}^{3} \sum_{k = 0}^{\infty}\pars{t^{4}}^{k}\,\dd t = {1 \over 6}\int_{0}^{1}{\pars{1 - t}^{3} \over 1 - t^{4}}\,\dd t \\[5mm] & = {1 \over 6} \int_{0}^{1}\pars{{2 \over 1 + t} - {1 \over 1 + t^{2}} - {t \over 1 + t^{2}}}\,\dd t = {1 \over 6}\bracks{2\ln\pars{2} - {\pi \over 4} - {1 \over 2}\,\ln\pars{2}} \\[5mm] & = \bbx{{1 \over 4}\,\ln\pars{2} - {1 \over 24}\,\pi} \approx 0.0424 \end{align}