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

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


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

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):

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

$\text{(3a)}$: 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




and, for completemess,


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.

  • $\begingroup$ @Mark Viola: I have tacitly assumed the common definition $H_0 = 0$. But thanks for the hint. I have incorporated this case now explicitly in the question. $\endgroup$ – Dr. Wolfgang Hintze Sep 9 '17 at 14:51
  • $\begingroup$ It might help if you showed us the first several values for $h_0$ and $h_1$. $\endgroup$ – Barry Cipra Sep 9 '17 at 14:58
  • $\begingroup$ @Barry Cipra Thank you for the valuable comment. Done. $\endgroup$ – Dr. Wolfgang Hintze Sep 9 '17 at 15:16

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

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 $\text{(1a)}$

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

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

  • $\begingroup$ Very interesting and nicely laid out - comme il faut! I have already filled lots of paper pages with this type of calculation. You might wish to consult the main OP referred to in the beginning, and especially the discussion on closed forms of finite sums. $\endgroup$ – Dr. Wolfgang Hintze Sep 21 '17 at 9:03

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