# more zeta splitting

$\phi=\frac{1}{2}(1+\sqrt{5})$

$\zeta(\phi)=\sum_{n=1}^{\infty} \frac{1}{n^{\phi}}$

therefore, amusingly:

$\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n+1}{n+1}\right)$

$\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n+1}{n+1}\right) = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n}{n+1}+\frac{1}{n+1}\right)$

$\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{n}{n+1}+\frac{1}{n+1}\right) = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{1}{1+n^{-1}}+\frac{1}{1+n}\right)$

$\sum_{n=1}^{\infty} \frac{1}{n^{\phi}} \left(\frac{1}{1+n^{-1}}+\frac{1}{1+n}\right) = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}+n^{\phi-1}} + \frac{1}{n^{\phi}+n^{\phi+1}}$

finally:

$\sum_{n=1}^{\infty} \frac{1}{n^{\phi}+n^{\phi-1}} + \frac{1}{n^{\phi}+n^{\phi+1}} = \sum_{n=1}^{\infty} \frac{1}{n^{\phi}+\sqrt[\phi]{n}} + \frac{1}{n^{\phi}+n^{\phi^{2}}}$

# splitting zeta

$\sum_{n=1}^{\infty}\left(\frac{1}{n^{2}+\sqrt{n}}\right)+\sum_{n=1}^{\infty}\left(\frac{1}{n^{2}+\sqrt{n^{7}}}\right)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\pi^{2}/6$

# infinite product of n=0 to infinity of (1+1/e^n)^(1/n!)

$A=\prod_{n=0}^{\infty} \sqrt[n!]{1+\frac{1}{e^{n}}}$

$\log A=\sum_{n=0}^{\infty} \frac{1}{n!} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}\left(\frac{1}{e^{n}}\right)^{m}$

$\log A=\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=0}^{\infty} \frac{(e^{-m})^{n}}{n!}$

$\exp\left[\sum_{m=1}^{\infty} \frac{(-1)^{m+1}e^{e^{-m}}}{m}\right]=\prod_{n=0}^{\infty} \sqrt[n!]{1+\frac{1}{e^{n}}}$

# infinite product of n=1 to infinity of n * sin(1/n)

$\prod_{n=1}^{\infty} n \sin(1/n)$

Since:

$A=\prod_{n=1}^{\infty} n \sin(1/n)=\prod_{n=1}^{\infty} \prod_{m=1}^{\infty} \left(1-\frac{1}{n^{2}m^{2}\pi^{2}}\right)$

$\log(A) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \log\left(1-\frac{1}{n^{2}m^{2}\pi^{2}}\right)$

$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}}{v} \left[-\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}$

$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v} \left[\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}$

$\sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v}\frac{1}{\pi^{2v}} \left(\sum_{n=1}^{\infty} \frac{1}{n^{2v}}\right) \left(\sum_{m=1}^{\infty} \frac{1}{m^{2v}}\right)$

$\sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v}\frac{\zeta(2v)^{2}}{\pi^{2v}}$

(which can easily be evaluated in wolfram alpha, and converges much faster than the original double product)

Therefore:

$\prod_{n=1}^{\infty} n \sin(1/n)\approx 0.755363388518573214063...$

# this blog is…

A math diary. Although I really like keeping a paper math diary, it’s helpful if it’s available for everyone to read.