Craig Mendoza

2022-06-27

I ran across a cool series I have been trying to chip away at.
$\sum _{k=1}^{\mathrm{\infty }}\frac{\zeta \left(2k+1\right)-1}{k+2}=\frac{-\gamma }{2}-6\mathrm{ln}\left(A\right)+\mathrm{ln}\left(2\right)+\frac{7}{6}\approx 0.0786\dots$

feaguelaBapzo

We have
$\begin{array}{rcl}& & \sum _{n=2}^{\mathrm{\infty }}\left(-{n}^{3}\mathrm{log}\left(1-\frac{1}{{n}^{2}}\right)-n-\frac{1}{2n}\right)\\ & =& \underset{N\to \mathrm{\infty }}{lim}\sum _{n=2}^{N}\left(-{n}^{3}\mathrm{log}\left(1-\frac{1}{{n}^{2}}\right)-n-\frac{1}{2n}\right)\\ & =& \underset{N\to \mathrm{\infty }}{lim}\left(\sum _{n=2}^{N}-{n}^{3}\mathrm{log}\left(1-\frac{1}{{n}^{2}}\right)-\left(\frac{{N}^{2}+N-2}{2}\right)-\left(\frac{\mathrm{log}\left(N\right)}{2}-\frac{1}{2}+\frac{\gamma }{2}+O\left(\frac{1}{N}\right)\right)\right)\\ & =& \underset{N\to \mathrm{\infty }}{lim}\left[\sum _{n=2}^{N}\left(2{n}^{3}\mathrm{log}\left(n\right)-{n}^{3}\mathrm{log}\left(n+1\right)-{n}^{3}\mathrm{log}\left(n-1\right)\right)-\left(\frac{{N}^{2}+N-2}{2}\right)-\left(\frac{\mathrm{log}\left(N\right)}{2}-\frac{1}{2}+\frac{\gamma }{2}\right)\right]\end{array}$
In the sum on the last line, we may gather together the coefficients of each logarithm (terms at the boundary of the sum are a little funny), giving
$\begin{array}{rcl}& & \underset{N\to \mathrm{\infty }}{lim}\left[\sum _{n=2}^{N}\left(-6n\mathrm{log}\left(n\right)\right)+\mathrm{log}\left(2\right)+\left({N}^{3}+3{N}^{2}+3N+1\right)\mathrm{log}\left(N\right)-{N}^{3}\mathrm{log}\left(N+1\right)-\left(\frac{{N}^{2}+N-2}{2}\right)-\left(\frac{\mathrm{log}\left(N\right)}{2}-\frac{1}{2}+\frac{\gamma }{2}\right)\right]\\ & =& \underset{N\to \mathrm{\infty }}{lim}\left[\sum _{n=2}^{N}\left(-6n\mathrm{log}\left(n\right)\right)+\mathrm{log}\left(2\right)-{N}^{3}\mathrm{log}\left(1+\frac{1}{N}\right)+\left(3{N}^{2}+3N+1\right)\mathrm{log}\left(N\right)-\left(\frac{{N}^{2}+N-2}{2}\right)-\left(\frac{\mathrm{log}\left(N\right)}{2}-\frac{1}{2}+\frac{\gamma }{2}\right)\right]\\ & =& \underset{N\to \mathrm{\infty }}{lim}\left[\sum _{n=2}^{N}\left(-6n\mathrm{log}\left(n\right)\right)+\mathrm{log}\left(2\right)-{N}^{3}\left(\frac{1}{N}-\frac{1}{2{N}^{2}}+\frac{1}{3{N}^{3}}+O\left(\frac{1}{{N}^{4}}\right)\right)+\left(3{N}^{2}+3N+1\right)\mathrm{log}\left(N\right)-\left(\frac{{N}^{2}+N-2}{2}\right)-\left(\frac{\mathrm{log}\left(N\right)}{2}-\frac{1}{2}+\frac{\gamma }{2}\right)\right]\\ & =& \underset{N\to \mathrm{\infty }}{lim}\left[\sum _{n=2}^{N}\left(-6n\mathrm{log}\left(n\right)\right)+\left(3{N}^{2}+3N+\frac{1}{2}\right)\mathrm{log}\left(N\right)+\left(\frac{-3}{2}{N}^{2}+\frac{7}{6}-\frac{\gamma }{2}+\mathrm{log}\left(2\right)\right)\right]\\ & =& -6\mathrm{log}\left(\underset{N\to \mathrm{\infty }}{lim}\left(\frac{\prod _{n=1}^{N}{n}^{n}}{{N}^{{N}^{2}/2+N/2+1/12}{e}^{-{N}^{2}/4}}\right)\right)+\frac{7}{6}-\frac{\gamma }{2}+\mathrm{log}\left(2\right)\\ & =& -6\mathrm{log}\left(A\right)+\frac{7}{6}-\frac{\gamma }{2}+\mathrm{log}\left(2\right)\end{array}$
Here, I am taking
$A=\underset{N\to \mathrm{\infty }}{lim}\frac{\prod _{n=1}^{N}{n}^{n}}{{N}^{{N}^{2}/2+N/2+1/12}{e}^{-{N}^{2}/4}}$
as the definition of the Glaisher-Kinkelin constant.

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