# Inequalities to give bounds on generalised harmonic numbers?

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)}

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

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)

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

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.

• 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. Commented Feb 28, 2020 at 21:30
• @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 $\frac{\pi ^2}{6}-\frac{C_1}{n} < H_n^{(2)} < \frac{\pi ^2}{6}-\frac{C_2}{n}$? Commented Feb 29, 2020 at 2:40
• 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. Commented Feb 29, 2020 at 8:50
• @RichardBurke-Ward It is fine as a comment. Commented Feb 29, 2020 at 10:55

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.

• You may also find interesting results in the paper doi.org/10.1007/s11139-014-9636-x
– Gary
Commented Mar 31, 2020 at 12:47
• @ 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. Commented Apr 3, 2020 at 9:13
• @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.
– Gary
Commented Apr 3, 2020 at 16:07

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 $$(15)$$

• 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 $H_n^{(k)}$.
– Gary
Commented Apr 7, 2020 at 13:59

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, $$k\ge 2$$)

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

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

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

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

• 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. Commented Apr 2, 2020 at 3:11
• (1) The generalization is in the formula (1). Please read the text again carefully. (2) I have corrected the inequalities. Thanks for your question. Commented Apr 3, 2020 at 4:45
• 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. Commented Apr 3, 2020 at 8:52
• @ 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 $n\ge 1$. Commented Apr 3, 2020 at 9:08

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.