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.

• For first, we may have an asymptotic expansion for the sum $$\sum_{n=1}^{N}\frac{\log n}{2n-1}$$ but not for the series that is clearly divergent. Second point: are you sure your asymptotic expansion is not missing a $C\log n$ term? – Jack D'Aurizio Apr 11 '15 at 14:09
• Thank you for your comment. I also was rather surprised by this, but it seems that the $\log n$ term is not present in the expansion. – Anatoly Apr 11 '15 at 14:23
• The value of the infinite sum diverges to $\infty$. Do you want an approximation of the partial sums? That's a different question. – Thomas Andrews Apr 11 '15 at 14:53
• @Thomas Andrews: Yes, I am interested in the partial sum. I edited the question accordingly. – Anatoly Apr 11 '15 at 15:09

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).

• Many thanks for your answer. It is very interesting to note that - differently from other similar logarithmic sums - the asymptotic expansion in this case seems to have no "closed form" in the strict sense. Whatever the solution you choose, either that involving zeta values or that reported in your answer, the evaluation of the remaining sum is still complicated. – Anatoly Apr 11 '15 at 19:52
• @Anatoly Note however that $\gamma_1$ itself does not have closed form, it is more of a notation. But indeed the asymptotics of some other logarithmic sums can be expressed in terms of this quantity, which does not seem to be the case for $\bar S_N$ (unless this sum has some hidden magic properties). – Start wearing purple Apr 11 '15 at 20:04

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

• It's dangerous to add up your infinite sum of $\frac{1}{2^{l+1}}o(1)$ to get an $o(1)$, which is implicit in your answer. It works here, but it takes some care to make sure it is right. – Thomas Andrews Apr 11 '15 at 15:33
• @ThomasAndrews Agreed. True. It works here though and can be shown with a little more analysis of the error term for $\sum_{n=1}^N \dfrac{\log(n)}{n^l}$. – Leg Apr 11 '15 at 15:39
• Thanks for your answer and for all comments. As I already noted in my comment to O.L., it is intriguing to observe that the asymptotic expansion for an apparently "simple" summation such as $\displaystyle \sum_{1}^N \frac{\log{n}}{2n-1}$ seems to be considerably more complex than the expansion of the classical summation $\displaystyle \sum_{1}^N \frac{\log{n}}{n}$. In particular, it seems that there is no way to express the remaining sum of the constant term in a strictly explicit notation, which was the main target of my question. – Anatoly Apr 11 '15 at 20:11
• Do we know the constant for $\sum\limits_{k=1}^n\frac{\log(n)}n\sim\frac12\log(n)^2+C$? – robjohn Apr 12 '15 at 14:06
• @robjohn The Stieltjes constants are defined to be these objects, if that is what you mean by "we know". – Leg Apr 12 '15 at 14:12

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.

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

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

$$\frac{\pi^2}{24}- \frac{1}{4}\, \ln^2\left( 2 \right) \sim 0.291.$$
• $\frac{\pi^2}{24}-\frac{\ln^2 2}{4} \approx 0.2911$. Which is very far from $0.2282$. – Start wearing purple Apr 11 '15 at 17:30