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Let $H_n$ be the $n$th harmonic number and $H_n^{(k)}$ be the $n$th harmonic number of order $k$ as follows:

$$H_n=\sum_{m=1}^{n}\frac{1}{m}$$ $$H_n^{(k)}=\sum_{m=1}^{n}\frac{1}{m^k}$$

There are several inequalities giving upper and lower bounds on $H_n$, such as this one found on MathWorld (eqn 14):

$$\frac{1}{2(n+1)}<H_n-\ln n-\gamma<\frac{1}{2n}$$

where $\gamma$ is the Euler-Mascheroni constant:

enter image description here

Are there any equivalent inequalities for $H_n^{(k)}$? And how does one arrive at them?

Heuristically, the following seems to hold, and offer nice tight bounds:

$$n^{-k} \left(-\frac{n}{k-1}+\gamma-\frac{k}{12 n}-\frac{1}{n^3}\right) +\zeta (k)<H_n^{(k)}<n^{-k} \left(-\frac{n}{k-1}+\gamma-\frac{k}{12 n}+\frac{1}{n^3}\right) +\zeta (k)$$

For example, this is a plot with $k=1.8$:

enter image description here

Is this inequality valid? And how do I prove it?

NOTE: This is a substantial revision of the original question, which was unclear - and since which, I have found the above potential bounds on my own. The bounty is for validation and proof.

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  • $\begingroup$ See also here: en.wikipedia.org/wiki/Riemann_zeta_function $\endgroup$ Commented Feb 28, 2020 at 17:47
  • $\begingroup$ Thanks @Michael. I'm aware of the zeta function, yes. But the zeta function is a sum to infinity. I'm after bounds on the growth of partial sums - $\sum_{n=1}^k \frac{1}{n^s}$ rather than $\sum_{n=1}^{\infty} \frac{1}{n^s}$ - so, generalised harmonic numbers rather than the zeta function. $\endgroup$ Commented Feb 28, 2020 at 21:30
  • $\begingroup$ @RichardBurke-Ward May you obtain inequalities from asymptote, for example, from $H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$, you try to find constants $C_1, C_2$ such that $\frac{\pi ^2}{6}-\frac{C_1}{n} < H_n^{(2)} < \frac{\pi ^2}{6}-\frac{C_2}{n}$? $\endgroup$
    – River Li
    Commented Feb 29, 2020 at 2:40
  • $\begingroup$ Hi @River. Thank you for this. Yes, I can work with that. If you post it as an answer, I can mark it as correct. $\endgroup$ Commented Feb 29, 2020 at 8:50
  • $\begingroup$ @RichardBurke-Ward It is fine as a comment. $\endgroup$
    – River Li
    Commented Feb 29, 2020 at 10:55

4 Answers 4

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Let $n\geq1$ and $k\geq 2$. By the result of this paper, it holds that \begin{align*} H_n^{(k)} = \zeta (k) & + n^{ - k} \left( - \frac{n}{{k - 1}} + \frac{1}{2} - \sum\limits_{m = 1}^{M - 1} \frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (k + 2m - 1)}}{{\Gamma (k)}}\frac{1}{{n^{2m - 1} }} \right. \\ & -\left. \theta _M (n,k)\frac{{B_{2M} }}{{(2M)!}}\frac{{\Gamma (k + 2M - 1)}}{{\Gamma (k)}}\frac{1}{{n^{2M - 1} }} \right), \end{align*} where $M\geq 1$, and $0<\theta _M (n,k)<1$ is an appropriate number. The $B_m$ are the Bernoulli numbers. In particular, with $M=2$, $$ H_n^{(k)} < \zeta (k) + n^{ - k} \left( { - \frac{n}{{k - 1}} + \frac{1}{2} - \frac{k}{{12}}\frac{1}{n} + \frac{{k(k + 1)(k + 2)}}{{720}}\frac{1}{{n^3 }}} \right) $$ and $$ H_n^{(k)} > \zeta (k) + n^{ - k} \left( { - \frac{n}{{k - 1}} + \frac{1}{2} - \frac{k}{{12}}\frac{1}{n}} \right). $$ Note that the constant must be $1/2$ and not $\gamma$. It is also seen that for sufficiently large values of $k$, your upper bound is not valid.

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  • $\begingroup$ You may also find interesting results in the paper doi.org/10.1007/s11139-014-9636-x $\endgroup$
    – Gary
    Commented Mar 31, 2020 at 12:47
  • $\begingroup$ @ Gary +1 Great answer! But let me mention that my derivation of (5) based on (1) is self contained and furthermore much simpler than that of the reference doi.org/10.1007/s11139-014-9636-x as it fits completely into the few lines of my answer. $\endgroup$ Commented Apr 3, 2020 at 9:13
  • $\begingroup$ @Dr.WolfgangHintze My only problem is that your (5) is a divergent alternating expansion so you cannot use theorems (e.g., Leibniz) on convergent series to justify properties of the error terms. $\endgroup$
    – Gary
    Commented Apr 3, 2020 at 16:07
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Extending this answer, we get $$ \sum_{k=1}^n\frac{1}{k^z}=\zeta(z)+\frac{1}{1-z}n^{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}+O\left(n^{-3-z}\right)\tag1 $$ Integrating a Riemann-Stieltjes Integral by parts, we get $$ \begin{align} \sum_{k=1}^n\frac1{k^z} &=\int_{1^-}^{n^+}\frac1{x^z}\,\mathrm{d}\lfloor x\rfloor\tag2\\ &=\int_1^n\frac1{x^z}\,\mathrm{d}x-\int_{1^-}^{n^+}\frac1{x^z}\,\mathrm{d}\!\left(\{x\}-\tfrac12\right)\tag3\\[6pt] &=\frac1{1-z}\left(n^{1-z}-1\right)+\frac12n^{-z}+\frac12 -\int_1^nzx^{-1-z}\left(\{x\}-\tfrac12\right)\mathrm{d}x\tag4\\ &=\frac1{1-z}\left(n^{1-z}-1\right)+\frac12\left(n^{-z}+1\right)-\frac{z}{12}\left(n^{-1-z}-1\right)\\ &-\int_1^nz(z+1)x^{-2-z}\left(\tfrac12\{x\}^2-\tfrac12\{x\}+\tfrac1{12}\right)\,\mathrm{d}x\tag5\\ &=\frac1{1-z}\left(n^{1-z}-1\right)+\frac12\left(n^{-z}+1\right)-\frac{z}{12}\left(n^{-1-z}-1\right)\\ &-\int_1^nz(z+1)(z+2)x^{-3-z}\left(\tfrac16\{x\}^3-\tfrac14\{x\}^2+\tfrac1{12}\{x\}\right)\mathrm{d}x\tag6\\ \end{align} $$ Comparing $(1)$ and $(6)$ as $n\to\infty$ for $\mathrm{Re}(z)\gt1$, we get $$ \begin{align} \zeta(z) &=\frac1{z-1}+\frac12+\frac{z}{12}\\ &-z(z+1)(z+2)\int_1^\infty x^{-3-z}\left(\tfrac16\{x\}^3-\tfrac14\{x\}^2+\tfrac1{12}\{x\}\right)\mathrm{d}x\tag7 \end{align} $$ which, by analytic continuation, holds for all $z\ne1$.

For $z\ge-3$, we have $$ 0\le\int_n^\infty x^{-3-z}\left(\tfrac16\{x\}^3-\tfrac14\{x\}^2+\tfrac1{12}\{x\}\right)\mathrm{d}x\le\frac{n^{-3-z}}{384}\tag8 $$ On each interval $[k,k+1]$, we can replace $x^{-3-z}$ by $x^{-3-z}-\frac12\left(k^{-3-z}+(k+1)^{-3-z}\right)$. This doesn't change the integral since $$ \int_k^{k+1}\left(\tfrac16\{x\}^3-\tfrac14\{x\}^2+\tfrac1{12}\{x\}\right)\mathrm{d}x=0\tag{8a} $$ Furthermore, $$ \left\|x^{-3-z}-\tfrac12\left(k^{-3-z}+(k+1)^{-3-z}\right)\right\|_{L^\infty[k,k+1]}=\tfrac12\left(k^{-3-z}-(k+1)^{-3-z}\right)\tag{8b} $$ and $$ \left\|\tfrac16\{x\}^3-\tfrac14\{x\}^2+\tfrac1{12}\{x\}\right\|_{L^1[k,k+1]}=\frac1{192}\tag{8c} $$ Summing the product of $\text{(8b)}$ and $\text{(8c)}$ for $k\ge n$ yields $(8)$.

We can combine $(6)$, $(7)$, and $(8)$ to get $$ \begin{align} \sum_{k=1}^n\frac{1}{k^z} &=\frac1{1-z}n^{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}\\ &+\zeta(z)+z(z+1)(z+2)\int_n^\infty x^{-3-z}\left(\tfrac16\{x\}^3-\tfrac14\{x\}^2+\tfrac1{12}\{x\}\right)\mathrm{d}x\tag9 \end{align} $$ Combining $(8)$ and $(9)$ gives $$ 0\le\sum_{k=1}^n\frac{1}{k^z}-\left(\zeta(z)+\frac{n^{1-z}}{1-z}+\frac{n^{-z}}2-\frac{z\,n^{-1-z}}{12}\right)\le\frac{z(z+1)(z+2)n^{-3-z}}{384}\tag{10} $$ Note that $(10)$ yields $\zeta(0)=-\frac12$, $\zeta(-1)=-\frac1{12}$, and $\zeta(-2)=0$.


Estimate for $\boldsymbol{k\ne1}$

Translating $(10)$ into the symbols from the question, we get $$ \bbox[5px,border:2px solid #C0A000]{0\le H_n^{(k)}-\left(\zeta(k)-\frac{n^{1-k}}{k-1}+\frac{n^{-k}}2-\frac{k\,n^{-1-k}}{12}\right)\le\frac{k(k+1)(k+2)n^{-3-k}}{384}}\tag{11} $$ The next term in the Euler-Maclaurin Sum Formula is $+\frac{k(k+1)(k+2)n^{-3-k}}{720}$, which is close to the middle of the range in $(11)$


Estimate for $\boldsymbol{k=1}$

We can take the limit as $z\to1$ of $(6)$, where $\frac{n^{1-z}-1}{1-z}\to\log(n)$, to get $$ \sum_{k=1}^n\frac1k =\log(n)+\frac1{2n}-\frac1{12n^2}+\frac7{12}-\int_1^n\frac{2\{x\}^3-3\{x\}^2+\{x\}}{2x^4}\,\mathrm{d}x\,\tag{12} $$ which gives the Euler-Mascheroni constant to be $$ \gamma=\frac7{12}-\int_1^\infty\frac{2\{x\}^3-3\{x\}^2+\{x\}}{2x^4}\,\mathrm{d}x\,\tag{13} $$ and the bounds $$ 0\le\sum_{k=1}^n\frac1k-\left(\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}\right)\le\int_n^\infty\frac{2\{x\}^3-3\{x\}^2+\{x\}}{2x^4}\,\mathrm{d}x\tag{14} $$ Estimating as in $(8)$, we get $$ \bbox[5px,border:2px solid #C0A000]{0\le H_n-\left(\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}\right)\le\frac1{64n^4}}\tag{15} $$ The next term in the Euler-Maclaurin Sum Formula is $+\frac1{120n^4}$, which is close to the middle of the range in $(15)$

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  • $\begingroup$ The error in absolute value is always at most the absolute value of the first omitted term and it has the same sign. This is true for both $H_n$ and $H_n^{(k)}$. $\endgroup$
    – Gary
    Commented Apr 7, 2020 at 13:59
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We can obtain bounds from the asymptotic expansions of $H_{n}^{(k)}$ which can be derived from this exact relation valid for $k\ge 2$

$$H_{n}^{(k)} = \zeta(k) + \frac{1}{(k-1)!} \left(-\frac{\partial }{\partial n}\right)^{k-1} H_{n}\tag{1}$$

where $\zeta(k)=\sum_{i=1}^{\infty}\frac{1}{i^k}$ is the Riemann zeta function.

$(1)$ can be easily derived from the well known representation, valid for $k \ge 1$

$$H_{n}^{(k)}=\sum_{m=1}^{\infty}\left(\frac{1}{m^k}-\frac{1}{(n+m)^k}\right)\tag{2}$$

which, for $k=1$ reads

$$H_{n}^{(1)}=H_{n} = \sum_{m=1}^{\infty}\left(\frac{1}{m}-\frac{1}{(n+m)}\right)\tag{3}$$

Inserting the asymptotic expansion of $H_{n}$

$$H_{n} \underset{n\to\infty}\simeq \log(n) +\gamma +\frac{1}{2n} -\frac{1}{12 n^2}+\frac{1}{120 n^4} \mp\ldots\tag{4}$$

we get

$$H_{n}^{(k)} \underset{n\to\infty}\simeq \zeta (k)+\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}-\frac{k}{12 n}+\frac{\binom{k+2}{3}}{120 n^3}\mp \ldots\right)\tag{5}$$

Taking more an more terms of the asymptotic expansion into account we can easily derive a chain of inequalities starting like this (notice that they are valid even for $n \ge 1$, and, of course, $k\ge 2$)

$$H_n^{(k)}-\zeta (k)>-\frac{1}{n^k}\frac{n}{(k-1)}\tag{6a}$$

$$H_n^{(k)}-\zeta (k)<\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}\right)\tag{6b}$$

$$H_n^{(k)}-\zeta (k)>\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}-\frac{k}{12n}\right)\tag{6c}$$

$$H_n^{(k)}-\zeta (k)<\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}-\frac{k}{12n}+\frac{k (k+1) (k+2)}{720 n^3}\right)\tag{6d}$$

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  • $\begingroup$ Hi @Dr. Wolfgang Hintze. I need two more elements of clarification before I can mark this as answered. (1) Can you explain how generalise from the case where $k=1$ (which trivially is the non-generalised harmonic number), and (2) your equations 6(a) and 6(b) suggest that $H_n^{(k)}<\frac{1}{2n^k}$ - but this implies $\zeta(k)<H_n^{(k)}$ which is clearly incorrect. Please clarify. $\endgroup$ Commented Apr 2, 2020 at 3:11
  • $\begingroup$ (1) The generalization is in the formula (1). Please read the text again carefully. (2) I have corrected the inequalities. Thanks for your question. $\endgroup$ Commented Apr 3, 2020 at 4:45
  • $\begingroup$ Thank you for the changes. I was confused because before, it said "for $k=1$..." Now it says "for $k\ge1$..." it makes sense. Marked as answered. $\endgroup$ Commented Apr 3, 2020 at 8:52
  • $\begingroup$ @ Richard Burke-Ward I'm really sorry for having caused confusion, and I'm grateful for your pointing out my errors. BTW it is $k\ge 2$ and $n\ge 1$. $\endgroup$ Commented Apr 3, 2020 at 9:08
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LEMMA (see [1]) Let $f(x)$ be a function with Taylor series in $(-a,a)$, $a\geq 1$. Let also its Taylor series converges absolutely in $1$. Then exists a constant $c=c(f)$ such that $$ \sum^{M}_{k=1}f\left(\frac{1}{k}\right)=\int^{M}_{1}f\left(\frac{1}{t}\right)dt+c(f)+O\left(\frac{1}{M}\right)\tag 1 $$
Moreover $$ c(f)=f(0)+f'(0)\gamma+\sum^{\infty}_{s=2}\frac{f^{(s)}(0)}{s!}\left(\zeta(s)-\frac{1}{s-1}\right). $$ where $\zeta(s)$ is Riemann's zeta function.

PROOF. See [1].

PROPOSITION. For a function $f$ as in lemma It holds the following usefull generalized expansion:
$$ \frac{1}{x}\sum^{x}_{k=1}f\left(\frac{x}{k}\right)-\int^{1}_{1/x}f\left(\frac{1}{t}\right)dt=\frac{f(0)}{x}+f'(0)\left(\gamma+\frac{1}{2x}-\frac{1}{12 x^2}\right)+ $$ $$ +\frac{c(f,x)}{x}+\frac{f(1)-f(0)-f'(0)}{2x}-\frac{f'(1)-f'(0)}{12x^2}+O\left(x^{-4}\right)\textrm{, as }x\rightarrow+\infty\tag 2 $$ where $$ c(f,x)=\sum^{\infty}_{s=2}\frac{f^{(s)}(0)x^s}{s!}\left(\zeta(s)-\frac{1}{s-1}\right) $$ which is generalization of LEMMA.

For to prove it one can use $$ \sum^{x}_{k=1}\frac{1}{k}=\log(x)+\gamma+\frac{1}{2x}-\frac{1}{12x^2}+O\left(x^{-4}\right)\textrm{, }x\rightarrow\infty\tag 3 $$ $$ \sum^{\infty}_{k=x+1}\frac{1}{k^s}=\frac{1}{(s-1)x^{s-1}}-\frac{1}{2x^s}+\frac{s}{12x^{s+1}}+O\left(x^{-s-3}\right)\textrm{, }x\rightarrow\infty\tag 4 $$ $$ \frac{1}{x}\int^{x}_{1}f\left(\frac{1}{t}\right)dt=\frac{x-1}{x}f(0)+f'(0)\log(x)+\frac{1}{x}\sum^{\infty}_{s=2}\frac{f^{(s)}(0)x^s}{s!(s-1)}-\sum^{\infty}_{s=2}\frac{f^{(s)}(0)}{s!(s-1)}\tag 5 $$ and change of variables $t\rightarrow tx$ $$ \int^{x}_{1}f\left(\frac{x}{t}\right)dt=x\int^{1}_{1/x}f\left(\frac{1}{t}\right)dt. $$

Your relation is Application of Proposition with $f(x)=x^k$.

Also $$ \frac{1}{x}\int^{x}_{1}f\left(\frac{1}{t}\right)dt=-\int^{1/x}_{0}f\left(\frac{1}{t}\right)dt+\int^{1}_{0}f\left(\frac{1}{t}\right)dt\tag 6 $$ Hence if $E(f,N)$ denotes the error terms of the Riemann approximation of integral $$ \int^{1}_{0}f\left(\frac{1}{t}\right)dt\tag 7 $$ with the usual rectangular method, then $$ E(f,N)=-\int^{1/N}_{0}f\left(\frac{1}{t}\right)dt+\frac{f(0)}{N}+f'(0)\left(\gamma+\frac{1}{2N}-\frac{1}{12N^2}\right)+ $$ $$ +\frac{c(f,N)}{N}+\frac{f(1)-f(0)-f'(0)}{2N}-\frac{f'(1)-f'(0)}{12N^2}+O\left(N^{-4}\right)\textrm{, }N\rightarrow+\infty.\tag 8 $$ Provited that (7) exists.

REFERENCES

[1]: Nikos Bagis, M.L. Glasser. 'Integrals and Series Resulting from Two Sampling Theorems'. Sampling Theory in Singnal and Image Processing., Sampling Publishing, Vol. 5, No. 1, 2006.

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