A closed form of the family of series $\sum _{k=1}^{\infty } \frac{\left(H_k\right){}^m-(\log (k)+\gamma )^m}{k}$ for $m\ge 1$

Question

Introduction

Inspired by the work of Olivier Oloa [1] and the question of Vladimir Reshetnikov in a comment I succeeded in calculating the closed form of the sum

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

for $m=3$.

The cases known (to me) are

$$s_1 = -\gamma _1+\frac{\pi ^2}{12}-\frac{\gamma ^2}{2}= 0.728694...\tag{2a}$$

$$s_2 = -2 \gamma \gamma _1-\gamma _2+\frac{5 \zeta (3)}{3}-\frac{2 \gamma ^3}{3}= 1.96897 ...\tag{2b}$$

$$s_3 =-3 \gamma ^2 \gamma _1-3 \gamma \gamma _2-\gamma _3+\frac{43 \pi ^4}{720}-\frac{3 \gamma ^4}{4}=5.82174 ...\tag{2c}$$

Here $\gamma$ is Euler's gamma, and $\gamma_k$ is Stieltjes gamma of order k.

Notice that $s_2$ was calculated in [1]. For a check of my method I calculated all three cases.

Inspecting the available cases $m=1, 2, 3$ I tentatively propose here a general formula for the closed form of $s_m$.

$$ s_{m}= -m \sum _{j=1}^{m-1} \gamma ^j \gamma _{m-j}+a_m \zeta (m+1)-\gamma _m-\frac{m }{m+1}\gamma ^{m+1}\tag{3}$$

Here the coefficients are

$$a_1=\frac{1}{2},\; a_2=\frac{5}{3},\; a_3=\frac{43}{8}\tag{4}$$

With only three terms of $a_k$ OEIS [2] returns of course too many choices to be able to identify the sequences. And what's more, I don't know if the true sequence is contained in OEIS at all.

Unfortunately, I could not extend my method to $m=4$.

Questions

1. Prove the formula for $s(3)$

2. Calculate $s(4)$ and verify and try to complete (3)

References

[1] A closed form of the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$

[2] https://oeis.org/

Answer 1

First answer

I regret to inform you that the conjectured $a_m$ isn't always a rational number. Good news is, the conjectured sum does indeed have a closed form in terms of Euler sums.

Define the harmonic zeta function $\mathcal{H}^m(s)=\sum_{k=1}^{\infty}\frac{H_k^m}{k^s}$. The general sum has the closed form

$$\begin{split}\sum_{n=1}^{\infty}\frac{H_n^m-(\log n+\gamma)^m}{n}=& -\frac{m}{m+1}\gamma^{m+1}+(-1)^{m+1}\frac{\zeta(m+1)}{m+1}+\sum_{k=0}^{m-2}\binom{m+1}{k+1}\frac{(-1)^{k+m}\mathcal{H}^{k+1}(m-k)}{m+1} \\ & -\sum_{k=1}^{m}\binom{m}{k}\gamma_k\gamma^{m-k}\end{split}$$

Where $\zeta(s)$ denotes the Riemann zeta function and $\gamma_n$ denotes the Stieltjes constants .

Examples

$$ \sum_{n=1}^{\infty}\frac{H_n-\log n-\gamma}{n}=\frac{\zeta(2)}{2}-\frac{\gamma^2}{2}-\gamma_1$$ $$ \sum_{n=1}^{\infty}\frac{H_n^2-(\log n+\gamma)^2}{n}=\frac{5\zeta(3)}{3}-\frac{2\gamma^3}{3}-2\gamma\gamma_1-\gamma_2$$ $$ \sum_{n=1}^{\infty}\frac{H_n^3-(\log n+\gamma)^3}{n}=-\frac{3}{4}\gamma^4+\frac{43}{8}\zeta(4)-3\gamma^2\gamma_1-3\gamma\gamma_2-\gamma_3$$ $$ \sum_{n=1}^{\infty}\frac{H_n^4-(\log n+\gamma)^4}{n}=-\frac{4}{5}\gamma^5+\frac{79}{5}\zeta(5)+3\zeta(2)\zeta(3)-4\gamma^3\gamma_1-6\gamma^2\gamma_2-4\gamma\gamma_3-\gamma_4$$

As you can see, the first values of $a_n$ was rational because the Euler sums $\mathcal{H}(2)$ and $\mathcal{H}(3),\mathcal{H}^2(2)$ could be written in terms of $\zeta(3)$ and $\zeta(4)$ respectively in terms of rational numbers.

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Edit 11/8/2024

Lemma 1 \begin{equation}H_n=\log(n)+\gamma +\frac{1}{2n}-\sum_{a=2}^k \frac{B_{a}}{an^a} +\int_n^{\infty} \frac{\tilde B_k(x)}{x^{k+1}}\, dx\end{equation} Where $B_n$ are the Bernoulli numbers and $\tilde B_n(x)$ is the $n$-th periodic Bernoulli polynomial.

Proposition 1 \begin{equation}\mathcal{H}^{m}({s})=\sum_{j=0}^m \binom{m}{j}\frac{\gamma^{m-j}\Gamma(j+1)}{(s-1)^{j+1}}+\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}{}_m\tilde\gamma_n(s-1)^n\end{equation} Where ${}_m\tilde\gamma_n$ is the corresponding coefficient. The above proposition can be proved by recursively applying Lemma 1

Lemma 2 $${}_m\tilde\gamma_n=\lim_{N\to \infty}\left[\sum_{k=1}^N \frac{H^m_k \log^n(k)}{k}-\sum_{j=0}^{m} \binom{m}{j}\gamma^{m-j}\frac{\log^{n+j+1}(N)}{n+j+1}\right]$$

At $N\to \infty$, the following asymptotic formula holds

Proposition 2$$\sum_{n=1}^N \frac{H_n^{m-1}}{n}\sim \frac{H_N^m}{m}+\frac{(-1)^m}{m}\zeta(m)+\sum_{k=1}^{m-2}\binom{m}{k}\frac{(-1)^{m+k}\mathcal{H}^{k}({m-k})}{m}$$

By utilizing the above formulas, we can find the constant term of $\mathcal H^m$ around $s=1$

Theorem $${}_m\tilde\gamma_0=\frac{\gamma^{m+1}}{m+1}+(-1)^{m+1}\frac{\zeta(m+1)}{m+1}+\sum_{k=0}^{m-2} \binom{m+1}{k+1} \frac{(-1)^{k+m} \mathcal{H}^{k+1}({m-k})}{m+1}$$

Finally, by writing $s_m$ as $\displaystyle \sum_{n=1}^{\infty}\frac{H_n^m-(\log n+\gamma)^m}{n^s}$ and expanding it around $s=1$, we get the closed form of $s_m$

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Edit 20/8/2024

I was being careful not to disclose full details of my derivation because this was in my paper, and I don't want something like this to happen. After consideration, I feel like answering a 6 years old question is probably more important. Anyways, here are the proofs

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Lemma 1 is well known, it is the Euler Macluarin expansion of harmonic numbers.

Proof of Proposition 1.

Consider multiplying $\frac{H_n}{n^s}$ on both sides of lemma 1 and sum both sides. $$\mathcal{H}(s)=-\zeta'(s)+\gamma \zeta(s)+\frac{1}{2}\zeta({s+1})-\sum_{a=2}^k \frac{B_{a}}{a} \zeta({a+s}) +\sum_{n\geq 1}\frac{1}{n^s}\int_n^{\infty} \frac{\tilde B_k(x)}{x^{k+1}}\, dx$$

By the Laurent expansion of the Riemann zeta function, we get \begin{equation}\mathcal{H}(s)=\frac{1}{(s-1)^2}+\frac{\gamma}{s-1}+O(1)\end{equation} Similarly, by multiplying $\frac{H_n^{m}}{n^s}$ on both sides of lemma 1 and summing both sides, we get \begin{equation}\mathcal{H}^{m+1}(s)=-{\mathcal{H}^{m}}'(s)+\mathcal{H}^m(s)+\cdots\end{equation}

We will prove Proposition 1 by induction. We already proved the case for $m=1$. $$ \begin{split}\text{RHS}& = -\frac{d}{ds}\sum_{j=0}^m \binom{m}{j}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j}+\gamma\sum_{j=0}^m \binom{m}{j}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j}+O(1) \\ & =\sum_{j=0}^m \binom{m}{j}\frac{\Gamma(j+2)}{(s-1)^{j+2}}\gamma^{m-j}+\sum_{j=0}^{m} \binom{m}{j}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+O(1) \\ & =\sum_{j=1}^{m+1} \binom{m}{j-1}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+\sum_{j=0}^{m} \binom{m}{j}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+O(1) \\ & =\sum_{j=0}^{m+1} \binom{m}{j-1}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+\sum_{j=0}^{m+1} \binom{m}{j}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+O(1) \\ & =\sum_{j=0}^{m+1}\underbrace{\left[ \binom{m}{j-1}+ \binom{m}{j}\right]}_{\binom{m+1}{j}}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+O(1) \\ & =\sum_{j=0}^{m+1}\binom{m+1}{j}\frac{\Gamma(j+1)}{(s-1)^{j+1}}\gamma^{m-j+1}+O(1) =\text{LHS} \end{split}$$ This proves Proposition 1

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Proof of Lemma 2. We have \begin{equation} \begin{split}{}_{m}\tilde\gamma_{n} & =\lim_{s\to 0^-}\frac{d^n}{ds^n}\left\{\mathcal{H}^{m}({1-s})-\sum_{j=0}^m \binom{m}{j}\frac{\gamma^{m-j}(-1)^{j+1}}{s^{j+1}}\Gamma(j+1)\right\} \\ & =\lim_{s\to 0^-}\left\{\sum_{k=0}^{\infty} \frac{H^m_k}{k^{1-s}}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\frac{\gamma^{m-j}(-1)^{n+j+1}}{s^{j+1+n}}(j+1)^{\overline{n}}\Gamma(j+1)\right\}\\ & =\lim_{s\to 0^-}\left\{\sum_{k=0}^{\infty} \frac{H^m_k}{k^{1-s}}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\gamma^{m-j}{\underbrace{\frac{(-1)^{n+j+1}}{s^{j+1+n}}\Gamma(j+n+1)}_{\textstyle \int_1^{\infty}\log^{j+n}(t)t^{s-1}\, dt }}\right\} \\ & =\lim_{s\to 0^-}\left\{\sum_{k=0}^{\infty} \frac{H^m_k}{k^{1-s}}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\gamma^{m-j}\int_1^{\infty}\log^{j+n}(t)t^{s-1}\, dt\right\} \\ & =\lim_{s\to 0^-}\lim_{N\to\infty}\left\{\sum_{k=0}^{N} \frac{H^m_k}{k^{1-s}}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\gamma^{m-j}\int_1^{N}\log^{j+n}(t)t^{s-1}\, dt\right\} \\ & =\lim_{N\to\infty}\lim_{s\to 0^-}\left\{\sum_{k=0}^{N} \frac{H^m_k}{k^{1-s}}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\gamma^{m-j}\int_1^{N}\log^{j+n}(t)t^{s-1}\, dt\right\} \\ & =\lim_{N\to\infty}\left\{\sum_{k=0}^{N} \frac{H^m_k}{k}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\gamma^{m-j}\int_1^{N}\log^{j+n}(t)\, \frac{dt}{t}\right\} \\ & =\lim_{N\to\infty}\left\{\sum_{k=0}^{N} \frac{H^m_k}{k}\log^n(k)-\sum_{j=0}^m \binom{m}{j}\gamma^{m-j}\frac{\log^{j+n+1}(N)}{j+n+1}\right\} \end{split} \end{equation}

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Proof of Proposition 2. I will use $a_n\approx b_n$ to denote that $\lim(a_n-b_n)=0$. By summation by parts, we have $$\begin{split}\sum_{n=1}^N \frac{H_n^m}{n} & =H_{N+1}^{m}H_N-\sum_{n=1}^N H_n \Delta H_n^m \\ &=H_{N+1}^{m+1}-\frac{H^m_{N+1}}{N+1}+\sum_{k=0}^{m-1}\binom{m}{k}(-1)^{m-k}\sum_{n=1}^N \frac{H^k_{n+1} H_n}{(n+1)^{m-k}}\\ & =H_{N+1}^{m+1}-\frac{H^m_{N+1}}{N+1}+\sum_{k=0}^{m-1}\binom{m}{k}(-1)^{m-k}\left[\sum_{n=1}^{N+1} \frac{H_n^{k+1}}{n^{m-k}}-\sum_{n=1}^{N+1} \frac{H_n^k}{n^{m-k+1}}\right] \\ & =H_{N+1}^{m+1}-\frac{H^m_{N+1}}{N+1}+\sum_{k=0}^{m-1}\binom{m}{k}(-1)^{m-k}\sum_{n=1}^{N+1} \frac{H_n^{k+1}}{n^{m-k}}-\sum_{k=0}^{m-1}\binom{m}{k}(-1)^{m-k}\sum_{n=1}^{N+1} \frac{H_n^k}{n^{m-k+1}}\\ & \approx H_{N+1}^{m+1}+(-m)\sum_{n=1}^N \frac{H_n^m}{n} +\sum_{k=0}^{m-2}\binom{m}{k}(-1)^{m-k}\mathcal{H}^{k+1}{(m-k)}-\sum_{k=0}^{m-1}\binom{m}{k}(-1)^{m-k}\mathcal{H}^{k}{(m-k+1)}\\ \implies (m+1)\sum_{n=1}^N \frac{H_n^m}{n} & \approx H_{N+1}^{m+1}+\sum_{k=0}^{m-2} \binom{m}{k}(-1)^{k+m}\mathcal{H}^{k+1}{(m-k)}-\sum_{k=-1}^{m-2}\binom{m}{k+1}(-1)^{k+m+1}\mathcal{H}^{k+1}{(m-k)} \\ & = H_{N+1}^{m+1}+(-1)^{m+1} \zeta(m+1)+\sum_{k=0}^{m-2}{\underbrace{\left[\binom{m}{k}+\binom{m}{k+1}\right]}_{\textstyle \binom{m+1}{k+1}}}(-1)^{k+m}\mathcal{H}^{k+1}{(m-k)}\\ \implies \sum_{n=1}^N \frac{H_n^m}{n} & = \frac{H_{N+1}^{m+1}}{m+1}+(-1)^{m+1}\frac{\zeta(m+1)}{m+1}+\sum_{k=0}^{m-2} \binom{m+1}{k+1} \frac{(-1)^{k+m} \mathcal{H}^{k+1}{(m-k)}}{m+1}\end{split}$$

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Using Proposition 2 on Lemma 2 by writing $\frac{H_{N+1}^{m+1}}{m+1}$ using the asymptotic expansion $H_n=\log(n)+\gamma+O(\frac{1}{n})$ gives theorem.

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Lastly, we write $s_m$ as $\sum_{n=1}^{\infty}\frac{H_n^m-(\log n+\gamma)^m}{n^s}$ and consider the Laurent expansion at $s=1$, binomial expand $(\log n+\gamma)^m$ and using theorem gives the desired closed form.

Answer 2

Before showing my solution of the specific problem, I would like to make some remarks on useful methods.

Methods

To begin with, I describe briefly some methods to solve the following basic problem:

Given a sequence of numbers $a_k, k=1,2,3,...$ we wish to calculate the infinite sum $s=\sum_{k=1}^\infty a_k$.

To my knowledge there are three important methods to tackle the problem. The first two of which have been employed frequently here. I haven't yet seen the third method, which I have used here.

1. Limit of partial sums

Maybe the most natural method as it employs just the definition of an infinite sum.

Defining the partial sum as $s_n = \sum_{k=1}^n a_k$ we have

$$s = \lim_{n\to \infty } \, s_n$$

2. Integral representation

If the summand can be written as

$$a_k = \int_{0}^1 w(x,k) \, dx$$

with some function $w(x,k)$ then, interchanging sum and integral, we get

$$s = \int_{0}^1 W(x) \, dx$$

where

$$W(x) = \sum_{k=1}^\infty w(k,x)$$

3. Generating function

Here we form a function $g(x)$ by multiplying the $a_k$ by $x^k$ ("wrapping") an summing up:

$$g(x) =\sum_{k=1}^\infty x^k a_k$$

This hopefully results in a known function, called the generating function (g.f) of the $a_k$.

Then we have

$$s = \lim_{x\to 1 } \, g(x)$$

We also obtain immediately an expression for the alternating sum defined as

$$s_a = \sum_{k=1}^\infty (-1)^k a_k$$

by

$$s_a = \lim_{x\to -1 } \, g(x)$$

Interchanging limits must be justified, of course. But here we skip this labour as we are just looking for a heuristic tool to find a closed expression. Once we have it, we can verify its validity numerically.

More specifically, the generating function method will be applied to cases where $a_k = u_k - v_k$ and where the sums of $u_k$ and $v_k$ separately can be divergent.

Despite of the separate divergence, in many cases their generating functions exist separately (the factor $x^k$ "forces" convergence) and we have

$$g(x) = g_u(x) - g_v(x)$$.

Answer 3

Second answer

I have came up with a simpiler proof that doesn't rely on studying the function $\sum_{n=1}^{\infty} \frac{H^m_n}{n^s}$. Only Proposition 2 from my last answer is needed. This answer should be easier to read.

Again, denote $a_n\approx b_n$ as $\lim_{n\to\infty}(a_n-b_n)=0$, we need the asymptotic $ \sum_{n=1}^N \frac{H^m_n}{n}\approx \frac{H_{N+1}^{m+1}}{m+1}+(-1)^{m+1}\frac{\zeta(m+1)}{m+1}+\sum_{k=0}^{m-2} \binom{m+1}{k+1} \frac{(-1)^{k+m} \mathcal{H}^{k+1}{(m-k)}}{m+1}$ proved in my previous answer.

\begin{equation}\begin{split}& \sum_{n=1}^{\infty}\frac{H_n^m-( \log n+\gamma)^m}{n}\\= &\lim_{N\to\infty}\left[\sum_{n=1}^N\frac{H^m_n}{n}-\sum_{n=1}^N\frac{(\log n+\gamma)^m}{n}\right]\\ =& \lim_{N\to\infty}\left[\sum_{n=1}^N\frac{H^m_n}{n}-\sum_{n=1}^N\frac{1}{n}\sum_{k=0}^m\binom{m}{k}\gamma^{m-k}\log^k(n)\right] \\ = & \lim_{N\to\infty}\Bigg[\sum_{n=1}^N\frac{H^m_n}{n}-\sum_{k=0}^m\binom{m}{k}\gamma^{m-k}{\underbrace{\sum_{n=1}^N\frac{1}{n}\log^k(n)}_{\gamma_k+\frac{\log^{k+1}(N)}{k+1}}}\Bigg]\\= &\lim_{N\to\infty} \Bigg[\frac{H_{N+1}^{m+1}}{m+1}+(-1)^{m+1}\frac{\zeta(m+1)}{m+1}+\sum_{k=0}^{m-2} \binom{m+1}{k+1} \frac{(-1)^{k+m} \mathcal{H}^{k+1}{(m-k)}}{m+1}\\ &-\sum_{k=0}^{m}\binom{m}{k}\gamma^{m-k}\gamma_k-\sum_{k=0}^{m}\binom{m}{k}\gamma^{m-k}\frac{\log^{k+1}(N)}{k+1}\Bigg] \end{split}\end{equation}

We can write the harmonic number like the following

$$\begin{split}\frac{H_{N+1}^{m+1}}{m+1}& \approx\frac{H_{N}^{m+1}}{m+1} \\ & \approx \frac{(\log N+\gamma)^{m+1}}{m+1} \\ & =\sum_{j=0}^{m+1} \binom{m+1}{j}\log^j(N)\frac{\gamma^{m+1-j}}{m+1} \\ & =\sum_{j=1}^{m+1}{\underbrace{\binom{m+1}{j}\frac{1}{m+1}}_{\textstyle \binom{m}{j-1}\frac{1}{j}}}\log^j(N){\gamma^{m+1-j}}+\frac{\gamma^{m+1}}{m+1}\\ & =\frac{\gamma^{m+1}}{m+1}+\sum_{j=0}^m \binom{m}{j} \frac{\log^{j+1}(N)}{j+1}\gamma^{m-j}\\ \end{split}$$

Putting this back to the original formula gives

$$\begin{split}\sum_{n=1}^{\infty}\frac{H_n^m-( \log n+\gamma)^m}{n}= &\frac{\gamma^{m+1}}{m+1}+(-1)^{m+1}\frac{\zeta(m+1)}{m+1}+\sum_{k=0}^{m-2} \binom{m+1}{k+1} \frac{(-1)^{k+m} \mathcal{H}^{k+1}{(m-k)}}{m+1} \\ & -\sum_{k=0}^{m}\binom{m}{k}\gamma^{m-k}\gamma_k\end{split}$$

Finally, pulling out the $k=0$ th term of the last sum establishes the closed form.

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I did not put the proof as an edit in my previous answer because the answer is too long already. And posting a different solution makes the answer easier to read. Please inform me promptly if I should combine two answers.