Trouble evaluating the sum involving logarithm 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!  

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.
