Friday, August 10, 2018

logarithms - Series involving log $sum_{n=1}^{infty} left( nlog left(frac{2n+1}{2n-1}right)-1right) = frac{1-log 2}{2}$



Does anybody know how to prove this series?



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




I arrived at this through Mathematica.



I tried writing $\log \left(\frac{2n+1}{2n-1}\right)$ as $\int_0^1 \frac{1}{\frac{2n-1}{2}+x}dx$ and $-\sum_{k=1}^\infty \frac{(-2)^k}{k(2n-1)^k}$ but none of them worked.


Answer



Note that



\begin{align*}
\sum_{n=1}^{\infty}\left(n \log \left(\frac{2n+1}{2n-1}\right) - 1 \right)
&=\lim_{N\to\infty} \sum_{n=1}^{N}\left(n \log \left(\frac{2n+1}{2n-1}\right) - 1 \right)\\

&=\lim_{N\to\infty} \log\left[ e^{-N} \prod_{n=1}^{N} \left(\frac{2n+1}{2n-1}\right)^{n} \right] \\
&=\lim_{N\to\infty} \log\left[ e^{-N} \frac{2^{N} N! (2N+1)^{N}}{(2N)!} \right].
\end{align*}



By Stirling's formula, it follows that



$$ e^{-N} \frac{2^{N} N! (2N+1)^{N}}{(2N)!} \sim \sqrt{\frac{e}{2}}. $$



This immediately yields the desired answer.


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