Cfrgtkky

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 $s$.

$$ H_{1/n} = frac{zeta(2)}{n}-frac{zeta(3)}{n^2}+frac{zeta(4)}{n^3}-frac{zeta(5)}{n^4}+ldots $$
hence
$$ sum_{ngeq 1}frac{(-1)^{n+1}}{n}H_{1/n} = sum_{kgeq 2}(-1)^k zeta(k)eta(k) $$
where the series appearing in the RHS has to be intended à-la-Cesàro. Since
$$ sum_{kgeq 2}(-1)^k (zeta(k)-1)=frac{1}{2},qquad sum_{kgeq 2}(-1)^k (eta(k)-1)=2log 2-frac{3}{2} $$
we have
$$ sum_{ngeq 1}frac{(-1)^{n+1}}{n}H_{1/n} = 2log 2-frac{1}{2}+sum_{kgeq 2}(-1)^k(zeta(k)-1)(eta(k)-1) $$
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, $frac{1}{2pi}int_{0}^{2pi}f(e^{itheta}),dtheta$, and numerically approximated in a efficient way. We also have that $(-1)^{d+1}$ is a multiplicative function, hence

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

$$ sum_{ngeq 1}frac{(-1)^{n+1}}{n}H_{1/n} = sum_{ngeq 1}frac{d(n)}{(n+1)(2n+1)}$$
can be written as
$$ sum_{ngeq 1}[x^{2n}]left[sum_{mgeq 1}frac{x^{2m}}{1-x^{2m}}right]cdot 2int_{0}^{1}x^{2n}(1-x),dx = 2int_{0}^{1}(1-x)sum_{mgeq 1}frac{x^{2m}}{1-x^{2m}},dx $$
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
$$ sum_{ngeq 1}(-1)^{n+1} H_{1/n} =zeta(2)log(2)+sum_{kgeq 2}(-1)^{k+1}zeta(k+1)eta(k)$$
since
$$zeta(k+1)eta(k)=(2-2^{1-k})sum_{ngeq 1}frac{sigma(n)}{n^{1+k}}, $$
$$sum_{kgeq 2}(-1)^{k+1}zeta(k+1)eta(k)=-sum_{ngeq 1}frac{(3n+1)sigma(n)}{n^2(1+n)(1+2n)} $$
and
$$ sum_{ngeq 1}sigma(n) x^n = sum_{mgeq 1}frac{x^m}{(1-x^m)^2}.$$
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 $sum_{ngeq 1}frac{(-1)^{n+1}}{n}H_{1/n}$ is related to the average order of $d(n)$ (tackled by Dirichlet's hyperbola method) and the approximated evaluation of $sum_{ngeq 1}(-1)^{n+1}H_{1/n}$ 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.

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 $lim_{ktoinfty}Phi left(-1,1,1+frac{1}{k}right) = Phi left(-1,1,1right)=log(2)$ 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 $N(s) simeq 0.638288$