Summing Over Fractional Harmonic Numbers

Question

I apologize if this has been asked before, but I was unable to find any answers suitable for my needs so far.

I have been studying the harmonic numbers over the last few days and I am unable to evaluate a particular sum involving fractional harmonic numbers. Here is a quick run-down of what I have tried so far:

Starting with $H_n = \sum_{k=1}^{n}\frac{1}{n}$ we get that (I actually find this very satisfying) \begin{align*} H_n &= \sum_{n=1}^{k} \int_{0}^{1}x^{k-1}dx \\ &= \int_{0}^{1}\frac{1-x^{n+1}}{1-x}dx\\ &= -n\int_{0}^{1}x^{n-1}\ln(1-x)dx\\ &= n\int_{0}^{1}x^n\frac{d}{dx}\text{Li}_2(x)dx \end{align*}

Which allows for a very simple method to evaluate the sum \begin{align*} \sum_{n=1}^{\infty}\frac{H_n}{n^p}&=\sum_{n=1}^{\infty}\frac{1}{n^{p-1}}\int_{0}^{1}x^n\frac{d}{dx}\text{Li}_2(x)dx\\ &=\int_{0}^{1}\text{Li}_{p-1}(x)\frac{d}{dx}\text{Li}_2(x)dx\\ &=(p+2)\zeta(p+1)-\sum_{n=1}^{p-2}\zeta(p-n)\zeta(n+1)\\ \end{align*}

Now this expression for $H_n$ we can evaluate $H_{1/n}$ for non-zero $n$, so I decided to try and evaluate $$\sum_{n=1}^{\infty}\frac{H_{\frac{1}{n}}}{n}$$

I defined the function $$\lambda_p(x)=\sum_{n=1}^{\infty}\frac{x^{\frac{1}{n}}}{n^p} $$ and we have $$\frac{d}{dx}\lambda_p(x)=\frac{\lambda_{p+1}(x)}{x}$$ and then I tried to evaluate the sum and got

\begin{align*} \sum_{n=1}^{\infty}\frac{H_{\frac{1}{n}}}{n} &= \int_{0}^{1}\lambda_2(x)\frac{d}{dx}\text{Li}_2(x)dx \\ &=\text{Li}_2(1)\lambda_2(1)-\int_{0}^{1}\text{Li}_2(x)\frac{d}{dx}\lambda_2(x)dx\\ &=\zeta(2)^2-\int_{0}^{1}\lambda_3(x)\frac{d}{dx}\text{Li}_3(x)dx\\ &= \sum_{n=2}^{\infty}(-1)^n\zeta(n)^2 \end{align*}

Now, the LHS converges whereas the RHS does not, and I realised that $\lambda_p(x)$ must not be uniformly convergent (I guess that I was reckless). Does anyone have an idea on how to evaluate this sum? I would be grateful for suggestions on how to proceed. Thanks

Answer 1

I just realised how to solve this, so for any who are curious, here is my solution.

We have that for any $|y|<1$, then $$H_y = \sum_{n=2}^{\infty}(-1)^n\zeta(n)y^{n-1}$$

so now the sum becomes \begin{align*} \sum_{n=1}^{\infty}\frac{H_{\frac{1}{n^m}}}{n^p}&=1+\sum_{n=2}^{\infty}\frac{1}{n^p}\sum_{k=2}^{\infty}(-1)^k\zeta(k)\frac{1}{n^{m(k-1)}}\\ &=1+\sum_{k=2}^{\infty}(-1)^k\zeta(k)\left(\zeta(p+m(k-1))-1\right) \end{align*} with $m=p=1$

[EDIT]

I got the series for $H_n$ through integration by parts, which gives

\begin{align*}H_n &= n\int_{0}^{1}x^n\frac{d}{dx}\text{Li}_2(x)dx\\ &= \sum_{k=2}^{p-1}(-1)^k\zeta(k)n^{k-1}+(-1)^pn^{p-1}\int_{0}^{1}x^n\frac{d}{dx}\text{Li}_p(x)dx\\ \end{align*} so for $|n|<1$, we can take the limit as p goes to infinity

Answer 2

We have: $$ H_{\frac{1}{n}}=\sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+\frac{1}{n}}\right) $$ hence $$ S=\sum_{n\geq 1}\frac{H_{1/n}}{n}=\sum_{n\geq 1}\sum_{m\geq 1}\left(\frac{1}{mn}-\frac{1}{mn+1}\right)=\color{red}{\sum_{r\geq 1}\frac{d(r)}{r(r+1)}} \tag{1}$$ where $\sum_{r\geq 1}\frac{d(r)}{r^s}=\zeta(s)^2$ also follows from Dirichlet's convolution. I do not think the RHS of $(1)$ has a nice closed form, but we may for sure apply Parseval's theorem to write $S$ as an integral, since $$ \sum_{n\geq 2}\zeta(n) z^n = -z H_{-z},\qquad \sum_{n\geq 2}(-1)^n \zeta(n) z^n = z H_{z}\tag{2}$$ so $$ S = \lim_{\lambda\to 0^+}\sum_{n\geq 2}(-1)^n \zeta(n)^2 e^{-\lambda n} = -\frac{1}{2\pi}\int_{-\pi}^{\pi}H_{-e^{i\theta}}\cdot H_{e^{-i\theta}}\,d\theta \tag{3}$$ or apply creative telescoping to accelerate the convergence of the series given by the RHS of $(1)$.
For instance, we may exploit $$\begin{eqnarray*} S &=& \sum_{r\geq 1}\frac{d(r)}{\left(r+\frac{1}{2}\right)^2}+\sum_{r\geq 1}\frac{d(r)}{r(r+1)(2r+1)^2}\\&=&\sum_{r\geq 1}\frac{d(r)}{\left(r+\frac{1}{2}\right)^2}+\frac{1}{4}\sum_{r\geq 1}\frac{d(r)}{\left(r+\frac{1}{2}\right)^4}+\ldots\end{eqnarray*}\tag{4}$$ but by going this way we lose the chance to evaluate the single terms by Dirichlet's convolution.
Still another way is to exploit $$ S = \frac{1}{2}+\sum_{n\geq 2}(-1)^n(\zeta(n)^2-1)$$

$$=\frac{1}{2}+\sum_{n\geq 2}\iint_{(0,+\infty)^2}\frac{1}{e^{2x}-e^{x}}\left(\frac{1}{e^y-1}+\frac{1}{e^{y}}\right)\sum_{n\geq 2}(-1)^n \frac{x^{n-1}y^{n-1}}{(n-1)!^2}\,dx\,dy$$ $$=\frac{1}{2}+\iint_{(0,+\infty)^2}\frac{1}{e^{2x}-e^{x}}\left(\frac{1}{e^y-1}+\frac{1}{e^{y}}\right)\cdot\left(1-J_0(2\sqrt{xy})\right)\,dx\,dy. \tag{5}$$