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I've seen a similar question to this on this site: Calculate sum of $\sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n}$.

However, this sum is a little different because the argument within the natural log function does not match the denominator and the sum starts at n = 2 instead of 1.

When I plugged this into Wolfram Alpha I get a complex number as a result:

$$ \sum_{n=2}^{\infty}(-1)^n \frac{\ln(n-1)}{n} ≈-0.0834085 - 4.53932×10^-14 i$$

I can't tell if this is a April fool's joke by Wolfram or if this answer is legitimate. How does a complex number arise when all terms in the sum are real?

Bernard
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user3760593
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    It says $10^{-14}$. It's pretty clear it's supposed to be $0$ but wolfram only gave an approximate answer. – mathworker21 Mar 31 '18 at 19:54
  • Oh i see, the imaginary part is quite small yes. Still a little weird that there is like an imaginary piece at all, even if it is really small... – user3760593 Mar 31 '18 at 19:56
  • Wolfram Alpha is not always correct. Here, it must be using some numerical methods (and using complex numbers somewhere) to find the sum, which results in a rounding error somewhere and the appearance of the $\epsilon\cdot i$ term (for $\epsilon \simeq 10^{-14}$, which hints to the fact you can should discard it). – Clement C. Mar 31 '18 at 19:56
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    @user3760593 see if you can adapt the solution with 16 upvotes (as of now) from the link you gave to the related problem. You might be able to. – mathworker21 Mar 31 '18 at 19:58
  • In $\texttt{Mathematica}$, you can use the $\texttt{Chop}$ function to 'cut' those pretty small numbers. – Felix Marin Apr 01 '18 at 20:53
  • @mathworker21 I tried using the technique in that solution, but I can't get it to work. Are you sure that method works for this problem? – user3760593 Apr 07 '18 at 01:36

1 Answers1

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$$S=\sum_{n\geq 1}(-1)^{n+1}\frac{\log n}{n+1} $$ is conditionally convergent by Leibniz' test. By Frullani's theorem $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx$, hence

$$ S = \int_{0}^{1}\frac{u-u^2+u^2\log(2)-\log(1+u)}{u^2 \log u}\,du\approx -0.0966 $$ and in terms of the derivatives of the $\eta(s)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} $ function we have $$ S = -\eta'(1)+\eta'(2)-\eta'(3)+\eta'(4)-\ldots $$ where $-\eta'(1)=\frac{1}{2}\log^2(2)-\gamma\log(2)$ and the RHS is an absolutely convergent series, since $\eta'(s)$ has an exponential decay on $(1,+\infty)$. By the relation between the $\zeta$ and $\eta$ functions we have

$$ S = \frac{1}{2}\log^2(2)-\gamma\log(2)+2\log(2)-2\log^2(2)+\sum_{n\geq 2}(-1)^n\left(1-\frac{2}{2^n}\right)\zeta'(n) $$ or $$ \boxed{S = -\frac{3}{2}\log^2(2)+(2-\gamma)\log(2)-\sum_{n\geq 1}\frac{\log(n)}{(n+1)(2n+1)}}$$ which allows an accurate numerical evaluation: $S\approx -0.096614934732226887$.

Jack D'Aurizio
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