Calculate the infinite sum $\sum_{1}^\infty \frac{\log{n}}{2n-1}$ I would like to calculate an asymptotic expansion for the following infinite sum:
$$\displaystyle \sum_{1}^N \frac{\log{n}}{2n-1}$$
when $N$ tends to $\infty$. I found that the asymptotic expansion for this partial sum is
$$ \displaystyle \frac{\log^2{N}}{4}+0.2282077...$$
and I would be interested in writing this constant term in an explicit way. By similarity with other sums of the same type, I believe that an explicit expression should probably include $\displaystyle \gamma$ and the first Stieltjes constant $\displaystyle \gamma_1$, but I was not able to find it.
 A: An incomplete answer but I hope it may clarify some things. 
Let us introduce
\begin{align*}
S_N&=\sum_{n=1}^N\frac{\ln n}{2n},\qquad 
\bar{S}_N=\sum_{n=1}^N\frac{\ln n}{2n-1}.
\end{align*}
Now make two observations:


*

*the sum $\displaystyle C_N:=\bar S_N-S_N=\sum_{n=1}^N\frac{\ln n}{2n\left(2n-1\right)}$ converges as $N\to \infty$.

*the asymptotics of $S_N$ is known:
$$S_N=\frac14\ln^2N+\frac12\gamma_1+o\left(1\right).$$
Thus the constant we are looking for is nothing but
$$\frac12\gamma_1+C_{\infty}=\frac12\gamma_1+\sum_{n=1}^{\infty}\frac{\ln n}{2n\left(2n-1\right)}.$$
However the evaluation of the remaining infinite sum looks complicated (yet much simpler than the sum involving zeta values from another answer).
A: You could use the Euler-Maclaurin formula. Alternatively, we have
\begin{align}
\sum_{n=1}^{N} \dfrac{\log(n)}{2n-1} & = \sum_{n=1}^{N} \dfrac{\log(n)}{2n} \cdot \dfrac1{1-\dfrac1{2n}} = \sum_{n=1}^{N} \dfrac{\log(n)}{2n} \sum_{l=0}^{\infty} \left(\dfrac1{2n}\right)^l\\
& = \sum_{n=1}^{N} \dfrac{\log(n)}{2n} + \sum_{l=1}^{\infty} \dfrac1{2^{l+1}} \underbrace{\sum_{n=1}^{N} \dfrac{\log(n)}{n^{l+1}}}_{-\zeta'(l)+o(1)}\\
& \sim \sum_{n=1}^{N} \dfrac{\log(n)}{2n} - \underbrace{\sum_{l=1}^{\infty} \dfrac{\zeta'(l+1)}{2^{l+1}}}_{\text{some constant}}
\end{align}
And we know the asymptotic expansion for $\displaystyle \sum_{n=1}^{N} \dfrac{\log(n)}n$.
A: Here is a general technique to do things from scratch. You can use the integral

$$ \int_{1}^{n} \frac{\ln(x)}{2x-1} = -\frac{1}{24}\,{\pi }^{2}-\frac{1}{2}\,\ln  \left( 2 \right) \ln  \left( 2\,n-1
 \right) - \frac{1}{2}\,{\it Li_2} \left( 1-2\,n \right) . $$

where $Li_s(z)$ is the polylogarith function. Note that you can use the asymptotic expansion for the function $Li_2(z)$ as

$$ Li_2(z) = -\frac{1}{2}\, \left( \ln  \left( z \right) +i\pi  \right) ^{2}- \frac{1}{6}\,{\pi }^{2
}-\frac{1}{z}-O(\frac{1}{z^2}).$$

Added: Here is your constant

$$ \frac{\pi^2}{24}- \frac{1}{4}\,  \ln^2\left( 2 \right)   
\sim 0.291. $$

