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# Tag Archives: Harmonic number

## Identities on the zeta function and harmonic numbers

Leonard Euler gave a general identity involving the harmonic number $H_n$ and the Riemann zeta function $\zeta(n)$ for natural numbers $n$.

$\displaystyle{2\sum_{n=1}^{\infty}\frac{H_n}{n^m} = (m+2)\zeta(m+1)-\sum_{k=1}^{m-2}\zeta(m-k)\zeta(k+1)}$.

A special case of this is the identity

$\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n^2} = 2\zeta(3)}$.

I first saw the above identity in Alex Youcis’s blog Abstract Nonsense and in course of further investigation, I was able to find several identities involving the Riemann zeta function and the harmonic numbers. While it is practically impossible to go through the entire mathematical literature to see if a formula is new or a rediscovery, in this post we shall see a few identities which are not given in the above links. Also if I encounter an identity elsewhere at any point of time, I have taken care to delete it from this post.

For $n \ge 1$ the following relations hold.

Entry 1.

$\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{2n}} - \frac{1}{(r+1)^{2n}}\Big\}=\zeta(2n+1)}$.

Entry 2.

$\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{4n-1}} - \frac{1}{(r+1)^{4n+1}}\Big\}=\zeta(4n)}$.

Entry 3.

$\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{4n+1}} - \frac{1}{(r+1)^{4n+1}}\Big\}=\zeta(4n+2)}$.

Entry 4.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{2n}} = n\zeta(2n+1) - \sum_{k=2}^{n}\zeta(k)\zeta(2n+1-k)}$.

Entry 5.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{4n-1}} = \frac{4n-3}{4}\zeta(4n) - \sum_{k=1}^{n-1}\zeta(2k+1)\zeta(4n-2k-1)}$.

Entry 6.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{4n+1}}= \frac{4n-1}{4}\zeta(4n+2)-\frac{\zeta(2n+1)^2}{2}}$

$\displaystyle{-\sum_{k=1}^{n-1}\zeta(2k+1)\zeta(4n-2k+1)}$.

Entry 7.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+2)^{2n}} = n\zeta(2n+1) - 2n - \sum_{k=2}^{n}\zeta(k)\zeta(2n+1-k) + \sum_{k=2}^{2n}\zeta(k)}$.

Entry a. If $|x| < 1$ then

$\displaystyle{\sum_{n=1}^{\infty}\zeta(4n)x^{4n} = \frac{1}{2}-\frac{\pi x}{4}(\cot(\pi x) +\coth(\pi x))}$.

## Zeta, gamma and the harmonic number

Leonard Euler discovered one the classical formula of analytical number theory;

$\displaystyle{\sum_{r=1}^{n} \frac{1}{r}=\ln n+\gamma+O\Big(\frac{1}{n}\Big)}.$

where $\gamma$ is the Euler–Mascheroni constant. The sum on the RHS is called the harmonic number and is denoted by $H_n$. In this article, I shall represent this classical asymptotic formula of Euler in terms of the Riemann zeta function.

Entry 1. If $|x| \gg 1$ then

$\displaystyle{\sum_{r=1}^{n} \zeta(r + 1/x) = n + x + \gamma + O(1/|x| + 1/n)}.$

Entry 2. Taking $x = -n$ we obtain a new formula for the Euler–Mascheroni constant

$\displaystyle{\lim_{n \rightarrow \infty}\sum_{r=1}^{n} \zeta(r - 1/n) = \gamma}.$

Entry 3. If $\{y\}$ denotes the fractional part of $y$ then

$\displaystyle{\sum_{r=1}^{n} \{\zeta(r + 1/n)\} = 1 + \gamma + O\Big(\frac{1}{n}\Big)}.$

Entry 4. If $|x| \gg 1$ then

$\displaystyle{\sum_{r=1}^{n} \frac{\zeta(r + 1/x)}{r} = \ln n + x + \gamma + O(1/|x| + 1/n)}.$

Entry 5. If $\{y\}$ denotes the fractional part of $y$ then

$\displaystyle{\sum_{r=1}^{n} \bigg \{\frac{\zeta(r + 1/n)}{r}\bigg \}=\ln n+\gamma+O\Big(\frac{1}{n}\Big)}.$

The RHS of the above asymptotic formula is exactly equal to that of Euler’s formula and thus we have

Entry 6.

$\displaystyle{\sum_{r=1}^{n} \bigg \{\frac{\zeta(r + 1/n)}{r}\bigg \}=H_n+O\Big(\frac{1}{n}\Big)}.$

## Zeta, gamma and the Stirling numbers

Ramanujan’s work on definite integrals is legendary. In the words of Hardy, “… he could of course evaluate any evaluable definite integral.” Today the great days of definite integrals and beautiful identities are long gone. Nevertheless Ramanujan’s work continues to inspire such beauties. In this article, I present a few definite integrals inspired by the works of Ramanujan. (I have not seen these above integral in mathematical literature. If anybody knows otherwise, please give me the reference.)

The integral formula

$\displaystyle{\int_{0}^{\infty} x^{t-1} e^{- ax} (\ln x)^r dx = \frac{\partial^r}{\partial t^r}\{a^{-t} \Gamma (t,a)\}}$

where $r=0,1,... , Re(a) >0, Re(t) > 0$ is well known and can be found in [1]. However this form does not help much in the sense that it does not make it any easier to evaluate the n-th order partial derivative in the RHS. In this article we shall consider definite integrals of the type

$\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- ax^m} (\ln x)^r dx}$

where $m$, $n$ and $r$ are positive integers and $a \ge 1$. We shall see that it is possible to explicitly represent evaluate such definite integrals in terms of the Euler–Mascheroni constant $\gamma$, the Riemann zeta function $\zeta (s)$ and the Stirling number of the first kind $s_{n,k}$. As usual let $H_n$ denote the harmonic number defined as $H_n = \sum_{r=1}^{n} \frac{1}{r}$. We define $I_n = \sum_{r=2}^{n} \frac{H_{r-1}}{r}$ and $J_n = \sum_{r=3}^{n} \frac{I_{r-1}}{r!}$. We have the following definite integrals.

Entry 1.

$\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- x^{m}} (\ln x)^3 dx}$

$\displaystyle{=\frac{(n-1)!}{m^4} \{-2\zeta(3) + 3\zeta(2)H_{n-1} - 3\gamma \zeta(2)}$

$\displaystyle{- 6\gamma I_{n-1}+ 3\gamma ^2 H_{n-1} - \gamma ^3 + 6J_{n-1}\}}$

$\displaystyle{=\frac{(-1)^n}{m^4}[\{2\zeta (3) + 3\gamma \zeta(2) - \gamma ^3\} s_{n,1}+ 3\{\zeta(2)+ \gamma ^2\} s_{n,2} + 6\gamma s_{n,3}+6 s_{n,4}]}$

Entry 2.

$\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- ax^{m}} (\ln x)^2 dx}$

$\displaystyle{=\frac{(n-1)!}{m^3 a^n}\{\zeta(2) - 2\gamma H_{n-1} + \gamma ^2 + 2I_{n-1}+ \ln^2 a - 2H_{n-1}\ln a + 2\gamma \ln a\}}$

$=\frac{(-1)^{n-1}}{m^4}[\{\zeta(2)+(\gamma + \ln a)^2 \}s_{n,1}+2(\gamma + \ln a)s_{n,2}+ 2s_{n,3}].$

Entry 3.

$\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- a x^{m}} \ln x dx = \frac{(n-1)!}{m^2 a^n} \{H_{n-1} - \gamma - \ln a\}.}$

Corollary. We have the following formula for the Stirling numbers of the first kind.

$\displaystyle{s_{n,2} = n!H_n} = n!\sum_{r=1}^{n}\frac{1}{r}$

$\displaystyle{s_{n,3} = -n!I_n} = -n!\sum_{r=2}^{n}\frac{H_{r-1}}{r}$

$\displaystyle{s_{n,4} = n!J_n} = n!\sum_{r=3}^{n}\frac{I_{r-1}}{r!}.$

References

[1] I.S. Gradshteyn and I.M. Ryzhik – Table of integrals series and products, (2007)