Calculate the infinite sum $\sum_{1}^\infty \frac{\log{n}}{2n-1}$

Question

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.

Answer 1

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

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

Answer 2

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 $\displaystyle \sum_{n=1}^{N} \dfrac{\log(n)}n$.

Answer 3

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).

Answer 4

Start with $$ \begin{align} \mathrm{Li}_2(x) &=-\int_0^x\frac{\log(1-t)}{t}\,\mathrm{d}t\\ &=\mathrm{Li}_2\left(\frac12\right)-\int_{1/2}^x\frac{\log(1-t)}{t}\,\mathrm{d}t\\ &=\mathrm{Li}_2\left(\frac12\right)+\int_{1-x}^{1/2}\frac{\log(t)}{t-1}\,\mathrm{d}t\\ &=\mathrm{Li}_2\left(\frac12\right)+2\int_{1/2-x/2}^{1/4}\frac{\log(2)+\log(t)}{2t-1}\,\mathrm{d}t\\ &=\mathrm{Li}_2\left(\frac12\right)-\log(2)\log(2x)+2\int_{1/2-x/2}^{1/4}\frac{\log(t)}{2t-1}\,\mathrm{d}t\tag{1} \end{align} $$ Substituting $x\mapsto1-2x$, we get $$ \mathrm{Li}_2(1-2x) =\mathrm{Li}_2\left(\frac12\right)-\log(2)\log(2-4x)+2\int_x^{1/4}\frac{\log(t)}{2t-1}\,\mathrm{d}t\tag{2} $$ which gives $$ \begin{align} \hskip{-6mm}\int_1^x\frac{\log(t)}{2t-1}\,\mathrm{d}t &=-\frac{\pi^2}{24}-\frac{\log(2)}2\log(2x-1)-\frac12\mathrm{Li}_2(1-2x)\\ &=\frac{\pi^2}{24}+\frac14\log(2x-1)^2-\frac{\log(2)}2\log(2x-1)+\frac12\mathrm{Li}_2\left(\frac1{1-2x}\right)\tag{3} \end{align} $$ where we have applied the Inversion Formula for $\mathrm{Li}_2$, proven in this answer.

Using the Euler-Maclaurin Sum Formula, we get $$ \begin{align} \hskip{-6mm}\sum_{k=1}^n\frac{\log(k)}{2k-1} &\sim C_1+\frac14\log(2n-1)^2-\frac{\log(2)}2\log(2n-1)+\frac12\mathrm{Li}_2\left(\frac1{1-2n}\right)\\ &+\frac12\frac{\log(n)}{2n-1}+\frac1{12}\left(\frac1{n(2n-1)}-\frac{2\log(n)}{(2n-1)^2}\right)\\ &-\frac1{720}\left(\frac2{n^3(2n-1)}+\frac6{n^2(2n-1)^2}+\frac{24}{n(2n-1)^3}-\frac{48\log(n)}{(2n-1)^4}\right)\tag{4} \end{align} $$ The asymptotic expansion in $(4)$ contains terms with up to $3$ derivatives of $\frac{\log(x)}{2x-1}$. Using the expansion containing terms with up to $11$ derivatives, and using $n=1000$, we can compute $$ C_1=0.348321017592010450605888035840979159864320\tag{5} $$ Combining $(5)$ and $$ \frac14\log(2n-1)^2-\frac{\log(2)}2\log(2n-1)=\frac{\log(n)^2}4-\frac{\log(2)^2}4+O\left(\frac{\log(n)}n\right)\tag{6} $$ we get your constant to be $C_2=C_1-\frac{\log(2)^2}4$ $$ C_2=0.228207764112460094439112404259312916931681\tag{7} $$ I have not yet found a closed form for $C_2$, but if one is found, we can use $(7)$ for confirmation.