Why mpmath computes $\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right)$

Question

Working with precision 500 decimal digits, mpmath in sage computes:

$$\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right).\tag{1}\label{1}$$

We believe the LHS of \eqref{1} diverges, so this isn't true.

Q1. Are there theoretical reasons for mpmath to compute \eqref{1}?

online code

Added Despite the interesting answers, I am ready bet mpmath doesn't do any analytic stuff not related to summation, it works purely numerically and the function is treated as black box, returning real number.

Answer 1

A summation method for this... $$ F(s) = -\sum_{n=2}^\infty \frac{(-1)^n}{n^s} = -(2^{1-s}-1)\zeta(s) \qquad\text{for $s>0$} $$ Differentiate: $$ \sum_{n=2}^\infty \frac{(-1)^n\log n}{n^s} = F'(s) = 2^{1-s}(\log 2)\zeta(s) - (2^{1-s}-1)\zeta'(s) \qquad\text{for $s>0$} $$

Now the term-by-term limit as $s \to 0^+$ would be $$ \sum_{n=2}^\infty (-1)^n\log n $$ which diverges, but $$ \lim_{s\to 0^+} \big(2^{1-s}(\log 2)\zeta(s) - (2^{1-s}-1)\zeta'(s)\big) = 2(\log 2)\zeta(0) -\zeta'(0) \\ = -\log 2 + \frac{\log(2\pi)}{2} = \frac{\log \pi - \log 2}{2} $$

Answer 2

As shown in my previous answer, the value of the sum that you see is $$\lim_{t\uparrow1}\sum_{n=2}^\infty (-t)^n \ln n. $$

Here is a "manual" way to show this. Writing \begin{equation} \ln n=\int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z, \end{equation} for $t\uparrow1$ we have \begin{equation} \begin{aligned} &\sum_{n=2}^\infty (-t)^n \ln n \\ &=\sum_{n=2}^\infty (-t)^n \int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z \\ &=\int_0^\infty \frac{dz}z\sum_{n=2}^\infty (-t)^n (e^{-z}-e^{-nz}) \\ &=\frac{t^2}{1+t}\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+t e^{-z}} \\ &\to\frac12\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+e^{-z}} \\ &=-\int_0^1 \frac{dx}{\ln x}\,\frac{1-x}{1+x} = \frac12\,\ln\frac\pi2, \end{aligned} \end{equation} by (say) Formula 4.267.1 on p. 545 in I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Seventh Edition, 2007.

Answer 3

The value of the sum that you see is

$$\lim_{s\downarrow-1}\sum_{n=2}^\infty s^n \ln n. $$

Here is the work in Mathematica:

Added by David Roberts (please see the corresponding comment of mine below): And here is the raw text for copy/paste, with output:

In[1]:=Sum[s^n Log[n], {n, 2, Infinity}, Assumptions->-1<s<1]

Out[1]=-s^2Derivative[0,1,0][LerchPhi][s,0,2]

In[2]:=Limit[%,s->-1,Direction->"FromAbove"]

Out[2]=-Derivative[0,1,0][LerchPhi][-1,0,2]

In[3]:=FullSimplify[-Derivative[0,1,0][LerchPhi][-1,0,2]]
Out[3]=(1/2)Log[Pi/2]