Is there a closed form for $\sum _{k=1}^n \frac{1}{k}H_{k-1}^2$?

Question

Introduction

This question appeared in the study (Sum of powers of Harmonic Numbers) of the sums of fourth powers of harmonic numbers $H_n = \sum _{k=1}^n \frac{1}{k}$ for $n=1,2,3,...$ and $H_0 = 0$.

Let the sum in question be

$$h_{0}(n) = \sum _{k=1}^n \frac{1}{k}H_{k-1}^2$$

Writing $H_{k-1} = H_{k} - 1/k$ we have

$$h_{0}(n) = h_{1}(n) - 2 h_{2}(n) + h_{3}(n)$$

Hence the question is equivalent to finding closed forms for

$$h_{1}(n)=\sum _{k=1}^n \frac{1}{k}H_{k}{}^2 \tag{1}$$

$$h_{2}(n) =\sum _{k=1}^n \frac{1}{k^{2}}H_{k} \tag{2}$$

$$h_{3}(n) =\sum _{k=1}^n \frac{1}{k^{3}}$$

Now $h_{3}$ is directly identified as a generalised harmonic number

$$h_{3} = H_n^{(3)}$$

Some Calculations

$h_{1}$ and $h_{2}$ are not independent as we shall prove next

$$h_{1}(n) - h_{2}(n) = \frac{1}{3} (H_n^3 - H_n^{(3)})\tag{3}$$

Let us first recall Abel's partial summation

$$\sum _{k=1}^n a_{k} b_{k}=A_{n} b_{k} + \sum _{k=1}^{n-1} A_{k} (b_{k}-b_{k+1})$$

where

$$A_{k}=\sum _{m=1}^k a_{m}$$

In order to caculate

$$h_{1}(n)=\sum _{k=1}^n \frac{H_k^2}{k}$$

we let $a(k)=\frac{1}{k},b(k)=H_k^2$ then $A_{k} = H_{k}$, and proceed with partial summation and (admittedly a little aimless) substitutions (borrowing the pretty layout from robjohn):

$$ \begin{align} \sum_{k=1}^n\frac{H_{k}^2}{k} &=H_n^3+\sum _{k=1}^{n-1} H_k \left(H_{k}^2-H_{k+1}^2\right)\tag{3a}\\ &=H_n^3+\sum _{k=1}^{n-1} H_k (H_k^2-\left(H_k+\frac{1}{k+1}\right)^2)\tag{3b}\\ &=H_{n}^3-\sum _{k=1}^{n-1} \frac{H_k \left(2 H_k\right)}{k+1}-\sum _{k=1}^{n-1} \frac{H_k}{(k+1)^2}\tag{3c}\\ &=H_n^3-2 \sum _{k=1}^{n-1} \frac{\left(H_{k+1}-\frac{1}{k+1}\right)^2}{k+1}-\sum _{k=1}^{n-1} \frac{H_{k+1}-\frac{1}{k+1}}{(k+1)^2}\tag{3d}\\ &=H_n^3-2 \sum _{k=1}^{n-1} \frac{H_{k+1}^2}{k+1}+4 \sum _{k=1}^{n-1} \frac{H_{k+1}}{(k+1)^2}-\sum _{k=1}^{n-1} \frac{H_{k+1}}{(k+1)^2}-2 \sum _{k=1}^{n-1} \frac{1}{(k+1)^3}+\sum _{k=1}^{n-1} \frac{1}{(k+1)^3}\tag{3e}\\ &=H_n^3-2 \sum _{k=1}^{n-1} \frac{H_{k+1}^2}{k+1}+3 \sum _{k=1}^{n-1} \frac{H_{k+1}}{(k+1)^2}-\sum _{k=1}^{n-1} \frac{1}{(k+1)^3}\tag{3f}\\ &=H_n^3-2 \sum _{m=2}^{n} \frac{H_{m}^2}{m}+3 \sum _{m=2}^{n} \frac{H_{m}}{m^2}-\sum _{m=2}^{n} \frac{1}{m^3}\tag{3g}\\ &=H_n^3-2 \sum _{k=1}^n \frac{H_k^2}{k}+3 \sum _{k=1}^n \frac{H_k}{k^2}-\sum _{k=1}^n \frac{1}{k^3}\tag{3h}\\[9pt] &3\sum_{k=1}^n\frac{H_{k}^2}{k}-3 \sum _{k=1}^n \frac{H_k}{k^2}= H_n^3-H_n^{(2)}\tag{3i} \end{align} $$

Explanation:
$\text{(3a)}$: partial summation
$\text{(3b)}$: use $H_{k+1}=H_k +\frac{1}{k+1}$
$\text{(3c)}$: expand bracket
$\text{(3d)}$: make $H_k$ compatible with the denominator using $H_{k}=H_{k+1} -\frac{1}{k+1}$
$\text{(3e)}$: expand terms
$\text{(3f)}$: collect terms
$\text{(3g)}$: substitute $k+1 \to m$
$\text{(3h)}$: start with $m=1$, as all first elements of the sums cancel, and finally let $m\to k$
$\text{(3i)}$: move the sums to the lhs. After division by $3$ the proof of (3) is completed.

I found (3) initially by just playing around with substitutions like in this derivation and stopped when I had found a nice formula.

We can also distribute the factors of the summand into the components of the partial summation differently, like for instance

$a(k)=\frac{H_k}{k},b(k)=H_k$ then $A_{k} = \sum_{m=1}^k\frac{H_{m}}{m} = \frac{1}{2} \left( H_k^2+H_k^{(2)}\right)$

This leads to another interesting formula

$$\sum _{k=1}^n \frac{H_k^2}{k}+\sum _{k=1}^n \frac{H_k^{(2)}}{k}=H_n H_n^{(2)}+\frac{1}{3}H_n^3+\frac{2}{3} H_n^{(3)}\tag{4}$$

Numerical series

The first several terms of the sequences are

$$h_{0}=\left\{0,\frac{1}{2},\frac{5}{4},\frac{301}{144},\frac{71}{24},\frac{82669}{21600},\frac{101191}{21600},\frac{23391199}{4233600},\frac{1074637}{169344},\frac{113452879}{15876000}\right\}$$

$$h_{1}=\left\{1,\frac{17}{8},\frac{701}{216},\frac{7483}{1728},\frac{1160603}{216000},\frac{1376693}{216000},\frac{543360959}{74088000},\frac{4894157017}{592704000},\frac{146372578939}{16003008000}\right\}$$

$$h_{2}=\left\{1,\frac{11}{8},\frac{341}{216},\frac{2953}{1728},\frac{388853}{216000},\frac{403553}{216000},\frac{142339079}{74088000},\frac{1163882707}{592704000},\frac{31983746689}{16003008000}\right\}$$

and, for completemess,

$$h_{3}=\left\{1,\frac{9}{8},\frac{251}{216},\frac{2035}{1728},\frac{256103}{216000},\frac{28567}{24000},\frac{9822481}{8232000},\frac{78708473}{65856000},\frac{19148110939}{16003008000}\right\}$$

The online-encyclopedia of integer sequences (https://oeis.org/) lists for numerator and denominator of the sequences the following

$$h_{3} = A007408/A007409$$ $$h_{2} = A195505/A195506$$

For $h_{1}$ only the denomintor is listed which, suprisingly, is the same as for $h_{2}$.

Notice also that the first 5 terms of the denominators of $h_{2}$ and $h_{3}$ are identical.

For $h_{0}$ neither numerator nor denominator are listed.

Answer 1

I don't have a closed form, but I derived an asymptotic approximation.

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Preliminaries $$ \begin{align} \sum_{k=1}^\infty\frac1{k^2}H_{k-1} &=\sum_{k=1}^\infty\frac1{k^2}\sum_{j=1}^{k-1}\frac1j\tag{1a}\\ &=\sum_{j=1}^\infty\frac1j\sum_{k=j+1}^\infty\frac1{k^2}\tag{1b}\\ &=\sum_{j=1}^\infty\frac1j\sum_{k=1}^\infty\frac1{(j+k)^2}\tag{1c}\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\left(\frac1{jk^2}-\frac1{k(j+k)^2}-\frac1{k^2(j+k)}\right)\tag{1d}\\ &=\frac12\sum_{j=1}^\infty\sum_{k=1}^\infty\left(\frac1{jk^2}-\frac1{k^2(j+k)}\right)\tag{1e}\\ &=\frac12\sum_{k=1}^\infty\frac1{k^2}H_k\tag{1f}\\ &=\frac12\left(\zeta(3)+\sum_{k=1}^\infty\frac1{k^2}H_{k-1}\right)\tag{1g}\\[9pt] &=\zeta(3)\tag{1h} \end{align} $$ Explanation:
$\text{(1a)}$: definition of Harmonic Numbers
$\text{(1b)}$: switch order of summation
$\text{(1c)}$: substitute $k\mapsto k+j$
$\text{(1d)}$: partial fractions
$\text{(1e)}$: average $\text{(1c)}$ and $\text{(1d)}$
$\text{(1f)}$: use analytic formulation for Harmonic Numbers
$\text{(1g)}$: $H_k=H_{k-1}+1/k$
$\text{(1h)}$: $2$ times $\text{(1g)}$ minus the left-hand side of $\text{(1a)}$

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Asymptotics $$ \begin{align} \hspace{-24pt}\sum_{k=1}^n\frac1kH_{k-1}^2 &=\sum_{k=1}^nH_{k-1}^2(H_k-H_{k-1})\tag{2a}\\ &=H_n^3+\sum_{k=1}^n\left(H_{k-1}^2-H_k^2\right)H_k\tag{2b}\\ &=H_n^3-\sum_{k=1}^n\frac1k\left(H_{k-1}+H_k\right)H_k\tag{2c}\\ &=H_n^3-\sum_{k=1}^n\frac1k\left(2H_{k-1}+\frac1k\right)\left(H_{k-1}+\frac1k\right)\tag{2d}\\ &=H_n^3-2\sum_{k=1}^n\frac1kH_{k-1}^2-3\sum_{k=1}^n\frac1{k^2}H_{k-1}-\sum_{k=1}^n\frac1{k^3}\tag{2e}\\ &=\frac13H_n^3-\sum_{k=1}^n\frac1{k^2}H_{k-1}-\frac13\sum_{k=1}^n\frac1{k^3}\tag{2f}\\[3pt] &=\frac13H_n^3-\frac43\zeta(3)+\frac{\log(n)}n+\frac{1+\gamma}n-\frac{\log(n)}{2n^2}-\frac{1+6\gamma}{12n^2}+O\!\left(\frac{\log(n)}{n^3}\right)\tag{2g} \end{align} $$ Explanation:
$\text{(2a)}$: $H_k-H_{k-1}=1/k$
$\text{(2b)}$: Summation by Parts
$\text{(2c)}$: $H_k-H_{k-1}=1/k$
$\text{(2d)}$: $H_k=H_{k-1}+1/k$
$\text{(2e)}$: expand the product
$\text{(2f)}$: add $\frac13$ of $\text{(2e)}$ to $\frac23$ of the left-hand side of $\text{(2a)}$
$\text{(2g)}$: apply $\text{(1h)}$ and the Euler-Maclaurin Sum Formula

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Limit $$ \lim_{n\to\infty}\left(\sum_{k=1}^n\frac1kH_{k-1}^2-\frac13H_n^3\right)=-\frac43\zeta(3)\tag3 $$