I was trying to solve this problem: Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
In the procedure I followed, I came across the following sum: $$\sum_{k=1}^{\infty} (-1)^{k-1}k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1}\right)$$
I cannot think of any approaches which would help me in evaluating the sum.
Any help is appreciated. Thanks!
The terms of the series vanish so we will look only at the partial sums where $n$ is odd.
$$
\begin{align}
&\sum_{k=1}^\infty(-1)^{k-1}k\left(\frac{\log(2k+1)}{2k+1}-\frac{\log(2k-1)}{2k-1}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^n(-1)^{k-1}k\frac{\log(2k+1)}{2k+1}-\sum_{k=0}^{n-1}(-1)^k(k+1)\frac{\log(2k+1)}{2k+1}\right)\tag{1}\\
&=\lim_{n\to\infty}\left((-1)^{n-1}n\frac{\log(2n+1)}{2n+1}+\sum_{k=1}^{n-1}(-1)^{k-1}\log(2k+1)\right)\tag{2}\\
&=\lim_{n\to\infty}\left((2n+1)\frac{\log(4n+3)}{4n+3}+\sum_{k=1}^{n}\log\left(\frac{4k-1}{4k+1}\right)\right)\tag{3}\\
&=\lim_{n\to\infty}\log\left((4n+3)^{\frac{2n+1}{4n+3}}\cdot\prod_{k=1}^{n}\frac{k-\frac14}{k+\frac14}\right)\tag{4}\\
&=\lim_{n\to\infty}\log\left(2\sqrt{n}\frac{\Gamma(n+\frac34)}{\Gamma(\frac34)}\frac{\Gamma(\frac54)}{\Gamma(n+\frac54)}\right)\tag{5}\\
&=\log\left(2\frac{\Gamma(\frac54)}{\Gamma(\frac34)}\right)\tag{6}\\
&=\log\left(\frac12\frac{\Gamma(\frac14)}{\Gamma(\frac34)}\right)\tag{7}\\
\end{align}
$$
Explanation:
$(1)$: write series as the limit of the partial sums, split into two sums, reindex the second sum
$(2)$: recombine the two sums
$(3)$: substitute $n\mapsto2n+1$ and combine pairs of terms of the sum
$(4)$: write a sum of logs as a log of a product
$(5)$: $\lim\limits_{n\to\infty}\frac1{2\sqrt{n}}(4n+3)^{\frac{2n+1}{4n+3}}=1$, write the product as a ratio of Gamma functions
$(6)$: Gautschi's inequality
$(7)$: $x\Gamma(x)=\Gamma(x+1)$
As said in comments, the sum oscillates between two values and converges to the value $0.391594$.
If we expand in the same spirit as Larsen, what we can see is that $$F(n)=\sum_{k=1}^{n} (-1)^{k-1}k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1}\right)$$ can be written as $$F(n)=\sum_{k=1}^{n-1} (-1)^{k-1}\log(2k+1)-\frac{(-1)^n n \log(2n+1)}{2n+1}$$
Added after AlexR's answer and comments
Let us consider the last term in AlexR' answer $$\Delta = (m+1) \left(\frac1{2m+3} \ln(2m+3) - \frac1{2m+1} \ln(2m+1)\right) $$ and let us expand it as a Taylor series for large values of $m$. We so obtain $$\Delta =\frac{\log \left(\frac{1}{m}\right)+1-\log (2)}{2 m}+\frac{-\log \left(\frac{1}{m}\right)-2+\log (2)}{2 m^2}+O\left(\left(\frac{1}{m}\right)^3\right)$$
On the limit
Experimenting a bit I find
$$\lim_{m\to\infty} S_m = \ln\left(\frac{\Gamma(\frac14)}{2\Gamma(\frac34)}\right) = 0.391594392706836...$$
To be the exact limit
On convergence
Let $a_k := \frac{\ln(2k-1)}{2k-1}$ then
$$\begin{align*}
S_m & = \sum_{k=1}^m (-1)^{k-1} k (a_{k+1} - a_k) \\
& = \sum_{k=2}^{m+1} (-1)^k (k-1) a_k + \sum_{k=1}^m (-1)^k k a_k \\
& \stackrel{a_1 = 0}= \sum_{k=2}^m (-1)^k ((k-1) a_k + k a_k) + (-1)^{m+1} ma_{m+1} \\
& = \sum_{k=2}^m (-1)^k \ln(2k-1) + (-1)^{m+1} \frac{m}{2m+1} \ln(2m+1)
\end{align*}$$
The problem all answerers overlooked is the factor of the last summand not being $\frac1{2m+1}$ but $\frac{m}{2m+1}$.
Let's first try to prove that the sequence is cauchy: $$\begin{align*} |S_{m+1}-S_m| & = |(-1)^{m+1} \ln(2m+1) \\ &\left. \qquad + (-1)^m \frac{m+1}{2m+3} \ln(2m+3) - (-1)^{m+1} \frac m{2m+1} \ln(2m+1) \right| \\ & = \left| (m+1) \left(\frac1{2m+3} \ln(2m+3) - \frac1{2m+1} \ln(2m+1)\right) \right| \\ & \le (m+1) \ln(2m+3) \left(\frac1{2m+1} - \frac1{2m+3}\right) \\ & = \ln(2m+3) (m+1) \frac2{4m(m+2) + 3} \\ & \le \ln(2m+3) \frac1{2m} \to 0 & (m\to\infty) \end{align*}$$
This proves convergence. The actual value seems a tad bit harder to prove.
