3
$\begingroup$

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 $s$.

$\endgroup$
3
$\begingroup$

$$ H_{1/n} = \frac{\zeta(2)}{n}-\frac{\zeta(3)}{n^2}+\frac{\zeta(4)}{n^3}-\frac{\zeta(5)}{n^4}+\ldots $$ 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.

$\endgroup$
0
$\begingroup$
  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.

  1. 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 $\zeta(2) \log(2) \simeq 1.14018$.

  1. 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 $N(s) \simeq 0.638288$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.