Closed expression for alternating harmonically wrapped harmonic series $\sum_{n=1}^\infty (-1)^{n+1} H_{\frac{1}{n}}$

Question

It is well known that the alternating harmonic sum $\sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n}$ converges to $\log(2)$.

Now let us wrap $\frac{1}{n}$ with the harmonic number $H_k$ (continued analytically to real values $k \to z$ as e.g. in https://math.stackexchange.com/a/3058569/198592) and ask if the sum

$$s = \sum_{n=1}^\infty (-1)^{n+1} H_{\frac{1}{n}}$$

is convergent.

Since for $n=1,2,3,...$ we have $H_{\frac{1}{n+1}}\lt H_{\frac{1}{n}}$ and $H_0=0$ the sequence $a_n = H_{\frac{1}{n}}$ is monotonous and goes to zero. Hence from the Leibniz criterion convergence is guaranteed.

The numerical value is

$$N(s) \simeq 0.638288$$

I wonder if there is a closed form expression for $s$.

Answer 1

$$ H_{1/n} = \frac{\zeta(2)}{n}-\frac{\zeta(3)}{n^2}+\frac{\zeta(4)}{n^3}-\frac{\zeta(5)}{n^4}+\ldots $$ hence $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}H_{1/n} = \sum_{k\geq 2}(-1)^k \zeta(k)\eta(k) $$ where the series appearing in the RHS has to be intended à-la-Cesàro. Since $$ \sum_{k\geq 2}(-1)^k (\zeta(k)-1)=\frac{1}{2},\qquad \sum_{k\geq 2}(-1)^k (\eta(k)-1)=2\log 2-\frac{3}{2} $$ we have $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}H_{1/n} = 2\log 2-\frac{1}{2}+\sum_{k\geq 2}(-1)^k(\zeta(k)-1)(\eta(k)-1) $$ in the classical sense. By exploiting the generating functions for $\zeta(k)-1$ and $\eta(k)-1$ the last series can be written as an inner product, $\frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta})\,d\theta$, and numerically approximated in a efficient way. We also have that $(-1)^{d+1}$ is a multiplicative function, hence

$$ \eta(k)\zeta(k)=\sum_{n\geq 1}\frac{1}{n^k}\sum_{d\mid n}(-1)^{d+1}=2\sum_{n\geq 1}\frac{d(n)}{n^k}-4\sum_{n\geq 1}\frac{d(n)}{(2n)^k} $$ and

$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}H_{1/n} = \sum_{n\geq 1}\frac{d(n)}{(n+1)(2n+1)}$$ can be written as $$ \sum_{n\geq 1}[x^{2n}]\left[\sum_{m\geq 1}\frac{x^{2m}}{1-x^{2m}}\right]\cdot 2\int_{0}^{1}x^{2n}(1-x)\,dx = 2\int_{0}^{1}(1-x)\sum_{m\geq 1}\frac{x^{2m}}{1-x^{2m}}\,dx $$ involving a classical Lambert series. The last integral function is given by the sum between $\log(1-x)$ and a continuous, bounded function on $[0,1]$, so the numerical approximation of the last integral is not difficult. The same techniques applies to $$ \sum_{n\geq 1}(-1)^{n+1} H_{1/n} =\zeta(2)\log(2)+\sum_{k\geq 2}(-1)^{k+1}\zeta(k+1)\eta(k)$$ since $$\zeta(k+1)\eta(k)=(2-2^{1-k})\sum_{n\geq 1}\frac{\sigma(n)}{n^{1+k}}, $$ $$\sum_{k\geq 2}(-1)^{k+1}\zeta(k+1)\eta(k)=-\sum_{n\geq 1}\frac{(3n+1)\sigma(n)}{n^2(1+n)(1+2n)} $$ and $$ \sum_{n\geq 1}\sigma(n) x^n = \sum_{m\geq 1}\frac{x^m}{(1-x^m)^2}.$$ For the pointwise evaluation of the involved Lambert series I suggest the approach through $\mathscr{M},\mathscr{M}^{-1}$ outlined here by Marko Riedel. It turns out that the approximated evaluation of $\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}H_{1/n}$ is related to the average order of $d(n)$ (tackled by Dirichlet's hyperbola method) and the approximated evaluation of $\sum_{n\geq 1}(-1)^{n+1}H_{1/n}$ is related to the average order of $\sigma(n)$. The magnitude of the error terms can be controlled through the Voronoi summation formula (about this, I can finally link the pages 126-144 of my notes); the optimal bounds are doomed to the depend on the shape of the zero-free region for the Riemann $\zeta$ function.

Answer 2

1. Jack D'Aurizio has given a contribution with many interesting relations to deep lying neighbouring fields.

He started from the expansion of the harmonic number $H_z$ around $z=0$.

This leads to a divergent sum

$$s_{JA} = \sum_{k=1}^\infty (-1)^{k+1} \eta(k)\zeta(k),\;\;\eta(k) = \left(1-2^{1-k}\right) \zeta (k) $$

whose partial sums oscillate between two finte values of order unity.

2. Here I propose another approach which avoids divergent series. It starts from the formula

$$H_{z} = \sum_{k=1}^\infty \frac{z}{k(k+z)}$$

leading after swapping the order of summation to

$$s_{WH} = \sum_{k=1}^\infty \frac{\Phi \left(-1,1,1+\frac{1}{k}\right)}{k^2}$$

where

$$\Phi(z,s,a) = \sum_{k=0}^\infty z^k(k+a)^{-s}$$

is the Lerch transcendent (http://mathworld.wolfram.com/LerchTranscendent.html).

The summands of $s_{WH}$ are positive and decreasing with $k$.

Since $\lim_{k\to\infty}\Phi \left(-1,1,1+\frac{1}{k}\right) = \Phi \left(-1,1,1\right)=\log(2)$ the sum is convergent and less than $\zeta(2) \log(2) \simeq 1.14018$.

3. We could also start from Euler's integral formula

$$H_{z} = \int_0^1 \frac{1-x^{z}}{1-x}\,dx$$

and replace the integrand letting $z \to \frac{1}{n}$

$$\frac{1-x^{1/n}}{1-x}$$

with its expansion about $x=1$ up to a given order. This then leads to a finite series of terms to be subtracted from $\log(2)$.

For example 10 terms give $s_{order 10} = 0.639872$ instead of the exact value $N(s) \simeq 0.638288$