fun-looking log sum. Seen this one before?.

[galactus]

Show that:

$$\sum_{n=1}^{\infty}\left(2n\log\left(\frac{4n+1}{4n-1}\right)-1\right)=\frac{1}{2}-\frac{1}{\pi}G-\frac{1}{4}\log(2)$$

G is a priori....the Catalan

There...some math Latin

[r9m]

Consider the $N$-th partial summation,

\begin{align*}S_N&=\sum\limits_{n=1}^{N} \left(2n\log \left(\frac{4n+1}{4n-1}\right) - 1\right) \\&= -N -\frac{1}{2}\sum\limits_{n=1}^{N} \log [(4n+1)(4n-1)] + \frac{1}{2}\sum\limits_{n=1}^{N} (4n+1)\log (4n+1) - (4n-1)\log (4n-1)\\&= -N - \frac{1}{2}\log \frac{(4N+1)!}{2^{2N}(2N)!}+\frac{1}{2}\sum\limits_{n=1}^{2N}(-1)^n (2n+1)\log (2n+1)\\&= -N + N\log 2 - \frac{1}{2}\log \frac{(4N+1)!}{2^{2N}(2N)!}+\sum\limits_{n=1}^{2N}(-1)^{n} \left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right)\\&\underbrace{\sim}_{\substack{N \to \infty\\\text{Stirling Approx.}}} - \frac{1}{2}\log \frac{2^{1/2}(4N+1)^{2N+1}}{e^{1/2}2^{2N}}+\sum\limits_{n=1}^{2N}(-1)^{n} \left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right)\\&= \frac{1}{4}-\frac{1}{4}\log 2 +\sum\limits_{n=0}^{2N}(-1)^{n} \left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right)-\left(N+\frac{1}{2}\right)\log \left(2N+\frac{1}{2}\right)\end{align*}

Also note the telescopic summation: $\displaystyle \frac{1}{2}\sum\limits_{n=0}^{2N} (-1)^n\left(n\log n + (n+1)\log (n+1)\right) = \left(N+\frac{1}{2}\right)\log (2N+1)$

Hence, \begin{align*}S_N &= \frac{1}{2} - \frac{1}{4}\log 2 + \sum\limits_{n=0}^{2N} (-1)^n\left(\left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right) - \frac{1}{2}n\log n - \frac{1}{2}(n+1)\log (n+1)\right) + o(1)\end{align*}

Now, $\displaystyle \lim\limits_{s \to 0} \frac{1}{s}\left(\frac{1}{\left(n+\frac{1}{2}\right)^{s-1}} - \frac{1}{2n^{s-1}}-\frac{1}{2(n+1)^{s-1}}\right) = -\left(\left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right) - \frac{1}{2}n\log n - \frac{1}{2}(n+1)\log (n+1)\right)$

as a result we have, \begin{align*}&\sum\limits_{n=0}^{2N} (-1)^n\left(\left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right) - \frac{1}{2}n\log n - \frac{1}{2}(n+1)\log (n+1)\right)\\&= \lim\limits_{s \to 0} \frac{1}{s}\sum\limits_{n=0}^{2N} (-1)^n\left(\frac{1}{2n^{s-1}}+\frac{1}{2(n+1)^{s-1}}-\frac{1}{\left(n+\frac{1}{2}\right)^{s-1}}\right)\\&= \lim\limits_{s \to 0} \frac{1}{s}\sum\limits_{n=0}^{2N} \frac{(-1)^n}{\Gamma(s-1)}\int_0^{\infty} \left(\frac{1}{2}+\frac{e^{-x}}{2} - e^{-x/2}\right)x^{s-2}e^{-nx}\,dx \\&= -\sum\limits_{n=0}^{2N} (-1)^n\int_0^{\infty} \left(\frac{1}{2}+\frac{e^{-x}}{2} - e^{-x/2}\right)x^{-2}e^{-nx}\,dx\end{align*}

Hence, $\displaystyle \lim\limits_{N \to \infty} S_N = \frac{1}{2}-\frac{1}{4}\log 2 - \frac{1}{2}\int_0^{\infty} \frac{1+e^{-x}-2e^{-x/2}}{x^2(1+e^{-x})}\,dx$

Now, to evaluate the above integral we use an indirect approach.

The following identity is well known (easy to prove): $\displaystyle \int_0^{\infty} \frac{t^{-s}s}{1+t}\,dt = \frac{\pi s}{\sin \pi s}$ and $\displaystyle \int_0^{1/2} \frac{\pi s}{\sin \pi s}\,ds = \frac{2G}{\pi}$

As a result we might write: \begin{align}\frac{2G}{\pi} &= \int_0^{\infty} \int_0^{1/2} \frac{t^{-s}s}{1+t}\,ds \,dt \\&= \int_0^{\infty} \frac{1- t^{-1/2} - \frac{1}{2}t^{-1/2}\log t}{(t+1)\log^2 t}\,dt\\&= \int_{-\infty}^{\infty} \frac{e^{-x}- e^{-x/2} + \frac{1}{2}xe^{-x/2}}{(1+e^{-x})x^2}\,dx \\&= \int_{0}^{\infty} \frac{e^{-x}- e^{-x/2} + \frac{1}{2}xe^{-x/2}}{(1+e^{-x})x^2}\,dx + \int_{-\infty}^{0} \frac{e^{-x}- e^{-x/2} + \frac{1}{2}xe^{-x/2}}{(1+e^{-x})x^2}\,dx \\&= \int_0^{\infty} \frac{1+e^{-x}-2e^{-x/2}}{x^2(1+e^{-x})}\,dx\end{align}

where, we made the substitution $t = e^{-x}$ in step $(2)$.

Combining the results, $$\sum\limits_{n=1}^{\infty} \left(2n\log \left(\frac{4n+1}{4n-1}\right) - 1\right) = \frac{1}{2}-\frac{1}{4}\log 2 -\frac{G}{\pi}$$

A much simpler solution due to Zaid Alyafeai can be found here