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


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 $$ S = \frac{1}{2}+\sum_{n\geq 2}(-1)^n(\zeta(n)^2-1)$$

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

  • $\begingroup$ That is very nice, thank you. I haven't heard of Parseval's theorem of Dirichlet's convolution, so I will surely look those up! I'm actually not too terribly concerned with closed form solutions, but more interested in how we can express sums (or generating functions) in terms of the poly logarithm. $\endgroup$ – Plopperzz Sep 29 '16 at 19:29

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 $m=p=1$


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


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