I have asked this question at math stack exchange, however it did not get any traction. Still curious about the answer though.

Numerical evidence suggests that:

$$\lim_{N \to +\infty} \sum_{n=1}^N\frac{1}{n}\sum_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}=s$$

or equivalently

$$\lim_{N \to +\infty} H\left( N \right)+\sum _{n=1}^{N} \left( { \frac {1}{n}\sum _{k=1}^{n}{\left( -1 \right) ^{k}{n\choose k}\frac { 1 }{ \left( k+1 \right) ^{s}}}} \right) = s$$

with $H(N)$ = the $N$-th Harmonic Number.

Convergence is quite slow, but clearly goes faster for negative $s$. Also, the computations for non-integer values of $s$ require high accuracy settings (I have used Maple, pari/gp and ARB).

However, according to Mathematica the series diverges by the "Harmonic series test", although when taking $s$ as an integer, it agrees on convergence.

Does this series converge for $s \in \mathbb{C}$ ?

Some numerical results below:

0.497702121, N = 100
0.499804053, N = 1000
0.499905919, N = 2000

-3.14160222, N = 100
-3.14159284, N = 1000
-3.14159272, N = 2000

2.45310498 - 1.94063637i, N = 100
2.33501943 - 2.09308517i, N = 1000
2.31996958 - 2.09923503i, N = 2000

1 Answer 1


We have to show that $$l(s):=\sum_{n=1}^\infty\frac1n\,S_n(s)=s,$$ where $$S_n(s):=\sum_{k=0}^n(-1)^k\binom nk\frac1{(k+1)^s} \\ =\sum_{k=0}^n(-1)^k\binom nk\int_0^\infty du\,u^{s-1}e^{-(k+1)u}/\Gamma(s) \\ =\int_0^\infty du\,u^{s-1}e^{-u}\sum_{k=0}^n(-e^{-u})^k\binom nk/\Gamma(s) \\ =\int_0^\infty du\,u^{s-1}e^{-u}(1-e^{-u})^n/\Gamma(s).$$ So, $$l(s)=\int_0^\infty du\,u^{s-1}e^{-u}\sum_{n=1}^\infty\frac1n\,(1-e^{-u})^n/\Gamma(s) \\ =\int_0^\infty du\,u^s e^{-u}/\Gamma(s) =\Gamma(s+1)/\Gamma(s) =s,$$ as desired.

  • $\begingroup$ very nice! So, the Mathematica divergence test wasn't correct after all. $\endgroup$
    – Agno
    Commented May 5, 2020 at 17:08

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