Posted on

# Ramanujan Formulae and Me(mes)

Choosing a page randomly and then saw…

$\large \sum_{k=0}^{\infty }\binom{2k}{k}\frac{1+2^{2k+1}}{2^{2k}5^{k+1}\left ( 2k+1 \right )^{2}}=\frac{\pi}{2\sqrt{5}}\log\left ( \frac{4}{2\sqrt{5}} \right )+\frac{1}{\sqrt{5}}\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}}{2^{2k+1}\left ( 2k+1 \right )^{2}}$

And finally…

$\large \pi=\frac{2\sqrt{5}}{\log\left ( \frac{4}{\sqrt{5}} \right )}\left ( \sum_{k=0}^{\infty }\binom{2k}{k}\frac{1+2^{2k+1}}{2^{2k}5^{k+1}\left ( 2k+1 \right )^{2}}-\frac{1}{\sqrt{5}}\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}}{2^{2k+1}\left ( 2k+1 \right )^{2}} \right )$

A beautiful yet complicated-enough formula for

But, what if I pushed it through to get anything useless like almost mathemacians would always certainly do?

And then I got…

$\large 2=\pi\left ( \log4-\log\sqrt{5} \right )\left (\sqrt{5} \sum_{k=0}^{\infty }\binom{2k}{k}\frac{1+2^{2k+1}}{2^{2k}5^{k+1}\left ( 2k+1 \right )^{2}}-\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}}{2^{2k+1}\left ( 2k+1 \right )^{2}} \right )^{-1}$

But, wait again! It has logarithms in it. So I could manipulate it algebraically to get an exponential form…

$\LARGE 2=e^{\left ( \frac{1}{4}\log5+\frac{\sqrt{5}}{\pi} \sum_{k=0}^{\infty }\binom{2k}{k}\frac{1+2^{2k+1}}{2^{2k}5^{k+1}\left ( 2k+1 \right )^{2}}-\frac{1}{\pi}\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}}{2^{2k+1}\left ( 2k+1 \right )^{2}} \right )}$

Forget it!!

Then I chose another page, again, randomly, and then saw another beautiful yet frightening formula…

$\large \sum_{k=0}^{\infty }\frac{2^{k}\left ( k! \right )^{2}}{\left ( 2k \right )!\left ( 2k+1 \right )^{2}}=-\frac{\pi}{2\sqrt{2}}\log\left ( 1+\sqrt{2} \right )-\frac{\pi^{2}}{4\sqrt{2}}+4\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}}{\left ( 4k+1 \right )^{2}}$

Hmmm… I coudn’t get a formula for  by that, but it has a logarithm in it. Might it be…???

$\LARGE 1=-\sqrt{2}+e^{\left ( -\frac{1}{2}\pi+\frac{8\sqrt{2}}{\pi}\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}}{\left ( 4k+1 \right )^{2}}-\frac{2\sqrt{2}}{\pi}\sum_{k=0}^{\infty }\frac{2^{k}\left ( k! \right )^{2}}{\left ( 2k \right )!\left ( 2k+1 \right )^{2}} \right )}$

And then I gave my results to Prof. Snape…

Prof. Snape: “By the way, you know what?

$\large e^{x}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}+\cdots$

Apply it to those formulae you’ve been invited…”

Seconds later…

.
.
.

===============

Betapa kurang kerjaan-nya gw di hari jum’at yang tenang ini…

#nowplaying Schubert – Ave Maria (again and again…)