# Closed form for the harmonic approximation sum $\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2$

Question

Is there a closed form of this harmonic approximation sum

$$s=\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2\tag{1}$$

The notation is standard.

Motivation

This question concerns a field which was treated frequently by many contributors here where, generally speaking, an appropriate functional of the difference of a summand an its asymptotic approximation is summed up, and mostly it is asked if there is a closed form of this sum in terms of "standard constants". The approximation as well as the functional must be chosen such that the resulting infinite sum in convergent. There are many examples of such "approximation sums".

Recently [1] I posted a question related to the approximation sum

$$\sum _{k=1}^{\infty } \left(H_k-(\log(k) + \gamma)\right){}^2\tag{2}$$

This led to the problem of a sum containing an uncommon $\log$-factor instead of the common polynomial in the index.

In an attempt to get rid of the $\log$ but still retain a non trivial problem I ask here to find a closed form for $s$ defined in (1).

A natural generalization is the generating function

$$s_{m,p}(x)=\sum _{k=1}^{\infty } x^k \left(H_k^{(m)}-\zeta (m)\right){}^p\tag{3}$$

To my knowledge, approximation sums containing modified harmonic numbers have not been investigated here.

The numerical value is

$$N(s_2) = 0.900362625200937377409205241520956358081230891307664$$

Solution attempt

We consider partial sums and write $s = \sigma_1+\sigma_2+\sigma_3$ where

$$\sigma_1 = \sum _{k=1}^{n } \left(H_k^{(2)}\right){}^2\tag{4a}$$ $$\sigma_2 =-2 \zeta(2) \sum _{k=1}^{n } H_k^{(2)}{}\tag{4b}$$ $$\sigma_3 = n \zeta(2)^2\tag{4c}$$

The sum in $\sigma_2$ can be calculated:

$$\sum _{k=1}^n H_k^{(2)}=(n+1) H_{n+1}^{(2)}-H_{n+1}\tag{5}$$

so that

$$\sigma_2 =-2 \zeta(2)\left( (n+1) H_{n+1}^{(2)}-H_{n+1}\right)\tag{6}$$

and we are left with $\sigma_1$ which can be easily shown by the reader to reduce to the calculation of either

$$h_2 = \sum _{k=1}^n \frac{H_k}{k^2}\tag{7}$$

or

$$h_4 = \sum _{k=1}^n \frac{H_{k}^{(2)}}{k}\tag{8}$$

These sums have been discussed (and christianed) earlier in [2] where also this relation has been derived

$$h_2+h_4 =H_n H_{n}^{(2)} + H_{n}^{(3)}\tag{9}$$

which means that knowledge of one of these function is suffient.

It would be nice to see an explicit form for the (finite) sums $h$ in terms of the elements of the set containing (modified) harmonic numbers and scalars.

But for the present task only the limit of large index must be determined.

References

• $(1)$ is not terribly hard to find by SBP, are you sure this wasn't asked before? – Jack D'Aurizio Jan 1 '18 at 18:12
• @JackD'Aurizio I'm not sure but I would appreciate relevant references. – Dr. Wolfgang Hintze Jan 1 '18 at 20:13
• I gave an answer below; mine was just a supposition. – Jack D'Aurizio Jan 1 '18 at 20:23

By summation by parts $$\begin{eqnarray*} -\sum_{k\geq 1}1\cdot\left(\zeta(2)-H_k^{(2)}\right)^2 &=& \sum_{n\geq 1}\frac{n}{(n+1)^4}+\frac{2n (H_n^{(2)}-\zeta(2))}{(n+1)^2}\\&=&\zeta(3)-\zeta(4)+2\sum_{n\geq 1}\frac{H_n^{(2)}-\zeta(2)}{n+1}-2\sum_{n\geq 1}\frac{H_n^{(2)}-\zeta(2)}{(n+1)^2}\\&=&\zeta(3)-\zeta(4)+2\zeta(2)-4\zeta(3)+\frac{7\pi^4-60\pi^2}{180}\end{eqnarray*}$$ (the last identity follows from standard Euler sums) and by simplifying we get $$\boxed{ \sum_{n\geq 1}\left(\zeta(2)-H_k^{(2)}\right)^2 = \color{red}{3\,\zeta(3)-\zeta(2)^2.}}$$ Similarly, $$\begin{eqnarray*}\sum_{k\geq 1}\left(\zeta(m)-H_k^{(m)}\right)^2 &=& \sum_{k\geq 1}\left[2\,\zeta(m)-H_{k}^{(m)}-H_{k+1}^{(m)}\right]\frac{k}{(k+1)^m}\\&=&\sum_{k\geq 1}\left[2\,\zeta(m)-2H_{k}^{(m)}-\frac{1}{(k+1)^m}\right]\cdot\left[\frac{1}{(k+1)^{m-1}}-\frac{1}{(k+1)^m}\right]\end{eqnarray*}$$ boils down to computing few values of $\zeta$, recalling $\sum_{k\geq 1}\frac{H_k^{(m)}}{k^m}=\frac{\zeta(2m)+\zeta(m)^2}{2}$ (symmetry) and tackling $\sum_{k\geq 1}\frac{H_k^{(m)}}{k^{m-1}}$ or $\sum_{k\geq 1}\frac{H_k^{(m-1)}}{k^m}$. Flajolet and Salvy outline efficient ways through residues. The first cases are $$\sum_{k\geq 1}\frac{H_k}{k^2}=2\,\zeta(3),\qquad \sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} = 3\,\zeta(2)\,\zeta(3)-\frac{9}{2}\,\zeta(5)$$ $$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^4} = -10\,\zeta(2)\,\zeta(5)+18\,\zeta(7).$$
• @ Jack D'Aurizio : Brilliant as always (I expected that, to be honest). The result is numerically ok, but could you please be a little more verbose with the derivation? It seems that my focusing on partial sums made things more complicated than necessary. And I have two questions: (a) what do you think of my question with respect to the finite sum (7) or (8)? (b) do you think that $s_{m,p}(1)$ leads to zeta values in general? And a final one: did find the time to you look at the problem of ref. [1] (with the log)? – Dr. Wolfgang Hintze Jan 1 '18 at 20:25
• Thank you but not necessary anymore. I see it now: partial summation with the factors $a_k = b_k = \zeta(2)-H_k^{(2)}$. – Dr. Wolfgang Hintze Jan 2 '18 at 6:20
• @Dr.WolfgangHintze: actually I applied summation by parts by writing $\left(\zeta(2)-H_k^{(2)}\right)^2$ as $1\cdot\left(\zeta(2)-H_k^{(2)}\right)^2$, then summing $1$ and "differentiating" $\left(\zeta(2)-H_k^{(2)}\right)^2$, but your approach is viable, too, and of course it leads to the same conclusion. – Jack D'Aurizio Jan 2 '18 at 17:48
• @ Jack D'Aurizio I just had finished the calculation in my self-answer when I saw that you had already treated this case $s_{m,2}(1). Hence we can compare the two solutions and procedures. And I have chosen the "verbose" manner (in the first place for myself as a reference). – Dr. Wolfgang Hintze Jan 2 '18 at 20:30 EDIT 03.01.18 • alternating sum$s_{m,2}(-1)$calculated up to the specific basic alternating sum$S^{+-}_{m,m} = \sum_{k=1}^\infty (-1)^k \frac{H_k^{(m)}}{k^m}$• brief discussion of$S^{+-}_{m,m}$. Closed forms are available for$m=1$and$m=2$in terms of$\zeta$-values,$\log(2)$, and Li. For$m \ge 3$reducibility remains to be clarified. • asymptotic behaviour of$s_{m,2}(1)$numerically determined Original post Inspired by the first part of the solution of Jack D'Aurizio I show here that the sum $$s_{m}=s_{m,2}(1) =\sum _{k=1}^{\infty } \left(\zeta (m)-H_k^{(m)}\right){}^2\tag{1}$$ has a closed form not only for$m=2$but generally for$m\ge 3$. Furthermore, this closed form is composed solely of zeta-values, i.e. of values of Riemann's$\zeta$function at positive integers$\ge 2$. Result Here are the first few expressions in the format$\{m, s_m\}$$$\begin{array}{c} \left\{2,- \zeta (2)^2+3\zeta (3)\right\} \\ \left\{3,-\zeta (3)^2+\pi ^2 \zeta (3)-10 \zeta (5)\right\} \\ \left\{4,-\frac{1}{3} 10 \pi ^2 \zeta (5)+35 \zeta (7)-\frac{\pi ^8}{8100}\right\} \\ \left\{5,-\zeta (5)^2+\frac{\pi ^4 \zeta (5)}{9}+\frac{35 \pi ^2 \zeta (7)}{3}-126 \zeta (9)\right\} \\ \left\{6,-\frac{1}{15} 7 \pi ^4 \zeta (7)-42 \pi ^2 \zeta (9)+462 \zeta (11)-\frac{\pi ^{12}}{893025}\right\} \\ \left\{7,-\zeta (7)^2+\frac{2 \pi ^6 \zeta (7)}{135}+\frac{28 \pi ^4 \zeta (9)}{15}+154 \pi ^2 \zeta (11)-1716 \zeta (13)\right\} \\ \left\{8,-\frac{1}{105} 8 \pi ^6 \zeta (9)-\frac{22 \pi ^4 \zeta (11)}{3}-572 \pi ^2 \zeta (13)+6435 \zeta (15)-\frac{\pi ^{16}}{89302500}\right\} \\ \left\{9,-\zeta (9)^2+\frac{\pi ^8 \zeta (9)}{525}+\frac{22 \pi ^6 \zeta (11)}{63}+\frac{143 \pi ^4 \zeta (13)}{5}+2145 \pi ^2 \zeta (15)-24310 \zeta (17)\right\} \\ \left\{10,-\frac{1}{945} 11 \pi ^8 \zeta (11)-\frac{286 \pi ^6 \zeta (13)}{189}-\frac{1001 \pi ^4 \zeta (15)}{9}-\frac{24310 \pi ^2 \zeta (17)}{3}+92378 \zeta (19)-\frac{\pi ^{20}}{8752538025}\right\} \\ \end{array}\tag{2a}$$ Or, in a "canonical" form, in which all powers of$\pi$are replaced by corresponding$\zeta$-functions $$\begin{array}{cc} 2 & - \zeta (2)^2+3\zeta (3)\\ 3 & -\zeta (3)^2+6 \zeta (2) \zeta (3)-10 \zeta (5) \\ 4 & -20 \zeta (2) \zeta (5)+35 \zeta (7)-\frac{7 \zeta (8)}{6} \\ 5 & -\zeta (5)^2+10 \zeta (4) \zeta (5)+70 \zeta (2) \zeta (7)-126 \zeta (9) \\ 6 & -42 \zeta (4) \zeta (7)-252 \zeta (2) \zeta (9)+462 \zeta (11)-\frac{715 \zeta (12)}{691} \\ 7 & -\zeta (7)^2+14 \zeta (6) \zeta (7)+168 \zeta (4) \zeta (9)+924 \zeta (2) \zeta (11)-1716 \zeta (13) \\ 8 & -72 \zeta (6) \zeta (9)-660 \zeta (4) \zeta (11)-3432 \zeta (2) \zeta (13)+6435 \zeta (15)-\frac{7293 \zeta (16)}{7234} \\ 9 & -\zeta (9)^2+18 \zeta (8) \zeta (9)+330 \zeta (6) \zeta (11)+2574 \zeta (4) \zeta (13)+12870 \zeta (2) \zeta (15)-24310 \zeta (17) \\ 10 & -110 \zeta (8) \zeta (11)-1430 \zeta (6) \zeta (13)-10010 \zeta (4) \zeta (15)-48620 \zeta (2) \zeta (17)+92378 \zeta (19)-\frac{524875 \zeta (20)}{523833} \\ \end{array}\tag{2b}$$ Asymptotic behaviour Numerically we find for large$m$a straight exponential decay: $$N(s_m)=1.01021 e^{-1.38663\; m}$$ Discussion Notice that for odd$m$the structure is simpler, at least the fraction is missing. The coefficient series are not in OEIS. Derivation We consider the partial sum up to$n$and in then take the limit$n\to\infty$. We shall use partial summation (PS) as was shown by Jack D'Aurizio in his solution to be the method of choice. Remember that partial summation (PS) allows to transform a sum similar to the well known partial integration (PI) $$\int_1^n A'(x) b(x) \, dx=A(x) b(x)|_1^n -\int_1^n A(x) b'(x) \, dx$$ namely $$\sum _{k=1}^n a_k b_k=A_n b_n- \sum _{k=1}^{n-1} A_k (b_{k+1}-b_k)\tag{3}$$ where$A_k=\sum _{j=1}^k a_j$is the "integral" of$a_k$, and$b_{k+1}-b_k$is the "derivative" of$b_k$. For a given summand the choice of the factors$a_k$and$b_k$is to a certain degreee arbitrary and one choice can be more favorable than the other. In any case the choice determines the rest of the calculations and hence should be mentioned in the beginning. Here we take $$a_k = b_k = \zeta (m)-H_k^{(m)}$$ Then we find For the "derivative" $$b_{k+1}-b_k=H_k^{(m)}-H_{k+1}^{(m)}=-\frac{1}{(1+k)^m}\tag{4a}$$ and for the "integral" $$A_k=\sum _{j=1}^k a(j)=H_{k+1}^{(m-1)}-(k+1) H_{k+1}^{(m)}+k \zeta (m)\tag{4b}$$ Here we have used the summation formula $$\sum _{k=1}^n H_k^{(m)}=(n+1) H_{n+1}^{(m)}-H_{n+1}^{(m-1)}\tag{5}$$ Plugging (4) into (3) the r.h.s. of (3) is given by the sum of four terms which are written down here already in the limiting form for$n \to \infty$$$R_1=A_n b_n \to 0\tag{6a}$$ $$R_2 = \sum _{k=1}^{\infty} \frac{k \zeta (m)}{(k+1)^m} \to \zeta (m) (\zeta (m-1)-\zeta (m))\tag{6b}$$ $$R_3 = -\sum _{k=1}^{\infty} \frac{H_k^{(m)}}{k^{m-1}}= - S_{m,m-1}\tag{6c}$$ $$R_4 = \sum _{k=1}^{\infty} \frac{H_k^{(m-1)}}{k^m}= S_{m-1,m}\tag{6d}$$ Here we have used the notation of [1] for$R_3$and$R_4$. These authors provide a formula of Borwein for$S_{p,q}$which is valid for the case of odd "weight"$w=p+q$. We are lucky here, as we have always odd weight (p+q=2m-1). Putting the parts together we have finally $$s_{m,2}(1) = \zeta (m) (\zeta (m-1)-\zeta (m)) - S_{m,m-1} + S_{m-1,m}\tag{7}$$ Some results have been shown above. They were checked numerically to be valid. In order to have everything in one place here is the function$S$$$S_{p,q}=(-1)^p \sum _{k=1}^{\left\lfloor \frac{q}{2}\right\rfloor } \zeta (2 k) \binom{-2 k+p+q-1}{p-1} \zeta (-2 k+p+q)+(-1)^p \sum _{k=1}^{\left\lfloor \frac{p}{2}\right\rfloor } \zeta (2 k) \binom{-2 k+p+q-1}{q-1} \zeta (-2 k+p+q)+\left(-\frac{1}{2} (-1)^p \binom{p+q-1}{p}-\frac{1}{2} (-1)^p \binom{p+q-1}{q}+\frac{1}{2}\right) \zeta (p+q)+\frac{1}{2} \left(1-(-1)^p\right) \zeta (p) \zeta (q)\text{/;}\text{OddQ}[p+q]$$ For the convenience of some users, here is also the Mathematica code for copying (not for reading) S[p_, q_] := (* = Sum[HarmonicNumber[k,p]/k^q,{k,1,\[Infinity]}], if p+q odd *) (Zeta[p + q] (1/2 - (-1)^p/2 Binomial[p + q - 1, p] - (-1)^p/2 Binomial[p + q - 1, q]) + (1 - (-1)^p)/2 Zeta[p] Zeta[q] + (-1)^p Sum[Binomial[p + q - 2 k - 1, q - 1] Zeta[2 k] Zeta[p + q - 2 k], {k, 1, Floor[p/2]}] + (-1)^p Sum[Binomial[p + q - 2 k - 1, p - 1] Zeta[2 k] Zeta[p + q - 2 k], {k, 1, Floor[q/2]}]) /; OddQ[p + q]  The alternating sum I have calculated the alternating sum $$s_a = s_{m,2}(-1) = \sum_{k\ge1} (-1)^k (\zeta(m) - H_k^{(m)})^2$$ using partial summation with the distribution$a_k = (-1)^k, b_k = (\zeta(m)-H_k^{(m)})^2$with the (preliminary?) result $$s_{m,2}(-1)=S^{+-}_{m,m}-\left(-2^{1-m} \zeta (m)^2-2^{-(2 m)} \zeta (2 m)+\frac{1}{2} \left(\zeta (m)^2+\zeta (2 m)\right)\right)\tag{8}$$ Where (using the definition of [1]) $$S^{+-}_{p,q} = \sum_{k=1}^\infty (-1)^k \frac{H_k^{(p)}}{k^q}$$ I have checked numerically that (8) is correct (to a high degree). The sum$S^{+-}_{p,q}$has been studied extensively in [1], but alas, closed forms are given (and seem to exist) only for odd weight$w = p + q$, and we have even weigth$2m$. Also in [2] our case, designated$\alpha_h(m,n)$there, is circumnavigated. For$m=1$the closed form result is known [3]: $$S^{+-}_{1,1} = \frac{\log ^2(2)}{2}-\frac{\zeta(2)}{2}$$ For$m=2$the question was asked in [4] and the answer was given by Przemo. Quoting also his notation it is $$S^{+-}_{2,2}={\bf H}^{(2)}_2(-1) = -4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)$$ It is not clear (to me) if his results really lead to closed form expressions for$m \ge 3$. Conclusion The question of a closed form was answered here affirmative for the non-alternating sum. For the alternating sum it could be traced back here to the reducibiity of the basic sum$S^{+-}_{m,m}$. It remains to be seen if the confirmed reducibility for$m=1$and$m=2$can be extended to higher$m$. References [1] http://algo.inria.fr/flajolet/Publications/FlSa98.pdf, Euler Sums and contour integral representations, Philippe Flajolet and Bruno Salvy, 1995 [2] http://www.davidhbailey.com/dhbpapers/eulsum-em.pdf, Experimental Evaluation of Euler Sums, David H. Bailey, Jonathan M. Borwein and Roland Girgensohn, 1994 [3]$S^{+-}_{1,1} = \frac{1}{12} \left(6 \log ^2(2)-\pi ^2\right)$, Proving an alternating Euler sum:$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$, asked by Spivey, answered by many [4]$S^{+-}_{2,2}$, Calculating alternating Euler sums of odd powers, asked by Zaid Alyafeai, answers were given by Przemo • This might be useful for tackling$s_{2,3}$. By summation by parts and a lot of patience, $$\sum_{k\geq 1}\frac{\left(H_k^{(2)}\right)^2}{k^2}=\frac{1}{3}\zeta(2)^3-\frac{2}{3}\zeta(6)+\zeta(3)^2.$$ – Jack D'Aurizio Jan 2 '18 at 22:44 • "Wunderwaffe" Jack once more. Congratulations! In the meantime I have calculated the alternating sum which, however, leaves another bone to crack:$\sum_{k\ge 1}(-1)^k \frac{H_k^{(m)}}{k^m}$. This is a case of even weigth for which [1] don't give an answer. – Dr. Wolfgang Hintze Jan 3 '18 at 18:04 I am adding a separate answer for showing a closed form for$s_{2,3}(1)$. By summmation by parts $$\sum_{k\geq 1}\left(\zeta(2)-H_m^{(2)}\right)^3=S_1+S_2+S_3$$ where $$S_1=\sum_{k\geq 1}\left[-\frac{1}{k^6}+\frac{1}{k^5}-\frac{\pi ^2}{2 k^4}+\frac{\pi ^2}{2 k^3}-\frac{\pi ^4}{12 k^2}+\frac{3 H_k^{(2)}}{k^4}-\frac{3 H_k^{(2)}}{k^3}+\frac{\pi ^2 H_k^{(2)}}{k^2}\right]$$ $$S_2=\sum_{k\geq 1}\left[\frac{\pi ^4}{12 k}-\frac{\pi ^2 H_{k}^{(2)}}{k}+\frac{3\left(H_k^{(2)}\right)^2}{k}\right]=3\sum_{k\geq 1}\frac{\left(\zeta(2)-H_k^{(2)}\right)^2}{k}$$ $$S_3 = -3\sum_{k\geq 1}\frac{\left(H_k^{(2)}\right)^2}{k^2}.$$$S_1$can be computed through standard Euler sums and$S_3$can be computed through the previous ones and summation by parts: $$S_1 = -2\zeta(6)-6\zeta(2)\zeta(3)+3\zeta(3)^2+\frac{29}{2}\zeta(5)$$ $$S_3 = 2\zeta(6)-3\zeta(3)^2-\zeta(2)^3$$ About$S_2$, summation by parts and Euler sums give $$S_2 = 21\zeta(2)\zeta(3)-21\zeta(5)-6\sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2}$$ and by Zaid's result outlined here we have $$\sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2}=\zeta(2)\,\zeta(3)+\zeta(5).$$ It follows that also$s_{2,3}$has a closed form in terms of values of the$\zeta\$ function: $$\boxed{ \sum_{k\geq 1}\left(\zeta(2)-H_{k}^{(2)}\right)^3 = \color{red}{9\,\zeta(2)\,\zeta(3)-\zeta(2)^3-\frac{25}{2}\zeta(5)}.}$$
\begin{align} S&=\sum_{n=1}^\infty\left(H_n^{(2)}-\zeta(2)\right)^2=\sum_{n=1}^\infty\left(-\int_0^1\frac{x^n\ln x}{1-x}\ dx\right)^2\\ &=\int_0^1\int_0^1\frac{xy\ln x\ln y}{(1-x)(1-y)}\sum_{n=1}^\infty(xy)^n\ dx=\int_0^1\int_0^1\frac{xy\ln x\ln y}{(1-x)(1-y)(1-xy)}\ dx\\ &=\int_0^1\frac{x\ln x}{1-x}\left(\int_0^1\frac{y\ln y}{(1-y)(1-xy)}\ dy\right)\ dx=\int_0^1\frac{x\ln x}{1-x}\left(\frac{\operatorname{Li}_2(x)-\zeta(2)x}{x(1-x)}\right)\ dx\\ &=\int_0^1\frac{\ln x}{(1-x)^2}\left(\operatorname{Li}_2(x)-\zeta(2)x\right)\ dx\quad \color{blue}{\text{apply IBP and simplify}}\\ &=\int_0^1\frac{\ln x\ln(1-x)}{x(1-x)}\ dx+\zeta(2)\int_0^1\frac{\ln x}{1-x}\ dx-\int_0^1\frac{\operatorname{Li}_2(x)-\zeta(2)x}{x(1-x)}\ dx\\ &=-\sum_{n=1}^\infty H_n\int_0^1x^{n-1}\ln x\ dx-\zeta^2(2)-\int_0^1\frac{\operatorname{Li}_2(x)-\zeta(2)x}{x}\ dx-\underbrace{\int_0^1\frac{\operatorname{Li}_2(x)-\zeta(2)x}{1-x}\ dx}_{IBP}\\ &=\sum_{n=1}^\infty\frac{H_n}{n^2}-\frac52\zeta(4)-\left(\operatorname{Li}_3(1)-\zeta(2)\right)-\left(-\int_0^1\frac{\ln^2(1-x)}{x}\ dx-\zeta(2)\int_0^1\ln(1-x)\ dx\right)\\ &=2\zeta(3)-\frac52\zeta(4)-\zeta(3)+\zeta(2)-\left(-2\zeta(3)-\zeta(2)(-1))\right)\\ &\boxed{=3\zeta(3)-\frac52\zeta(4)} \end{align}