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# Tag Archives: Riemann zeta function

## 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

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

## On the real roots of ζ(x) = a

Let $\mathbb N$ be the set of natural numbers and $\mathbb Q^+$ be the set of positive rational numbers. For every positive integer $n \ge 2$, there exists a unique real number $s_n$ such that $\zeta (s_n) = n$. The question is does there exists an $n \in \mathbb N$ such that $s_n \in \mathbb Q^+$ be rational? In other words, does there exist a positive rational $\frac{p}{q}$ such that $\zeta(\frac{p}{q}) \in \mathbb N$? At present, we don’t even know if $s_n$ is always rational, let alone irrational or transcendental. Investigation seems to suggest that $\zeta(\frac{p}{q}) \notin \mathbb N$ for any $\frac{p}{q} \in \mathbb Q^+$. However this has not been completely established. In this post, we shall look at some on the results in this line of research.

Expanding the Riemann zeta function about $s=1$ gives

$\displaystyle{\zeta(s) = \frac{1}{s-1} + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n (s-1)^n}$

where the constants

$\displaystyle{\gamma_n = \lim_{m \rightarrow \infty}\Bigg [\sum_{k=1}^{m}\frac{(\ln k)^n}{k} - \frac{(\ln m)^{n+1}}{n+1}\Bigg]}$

are known as Stieltjes constants. Inverting the above series we obtain:

Theorem 1Let $x > 1$ and $\zeta(x) = a$; then,

$\displaystyle{x = 1 + \frac{1}{a-\gamma _0} - \frac{\gamma _1}{1!(a-\gamma _0)^2} + \frac{\gamma _2}{2!(a-\gamma _0)^3}- \frac{\gamma _3 - 12\gamma _1}{3!(a-\gamma _0)^4} + O\Big(\frac{1}{a^5}\Big)}$.

Proof. The proof follows by inverting the Stieltjes series expansion of $\zeta (x)$$\Box$

We can rewrite the Stieltjes series expansion of the  Riemann zeta function in the following form which we shall use regularly in our analysis:

$\displaystyle{\zeta \Big (1+\frac{1}{s}\Big ) = s + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n\gamma_n}{n!^n}}$.

We want to find an interval where the function $\zeta(1+\frac{1}{s})$ is strictly decreasing. Clearly $\zeta(1+\frac{1}{s})$ is continuous and strictly decreasing in the interval $(0, \infty)$. The trivial zeros of $\zeta(s)$ occur at $s=-2n$ where $n \in N$. In other words, the trivial zeros of $\zeta(1+\frac{1}{s})$ occur at $s = \frac{-1}{2n+1}$ and all these zeros lie in the interval $(-1/3, 0)$. Hence in this interval, $\zeta(1+\frac{1}{s})$ will have infinitely many points of inflexion. Excluding the interval $(-1/3, 0)$, we can divide the real line in to two parts $(-\infty, -1/3)$ and $(0,\infty)$; and in each of these two intervals, $\zeta(1+\frac{1}{s})$ in strictly decreasing. For positive rationals $\frac{p}{q}$, the scope of our current study is the interval  $(0,\infty)$.

Lemma 2For every finite real $a>1$, there exists a unique real number $c_a, 0< c_a < 1-\gamma _0$, such that

$\displaystyle{\zeta\Big(1+\frac{1}{a-1 +c_a}\Big) = a}$.

Proof. Let lemma follows form the fact that in our interval of interest $(0,\infty)$$\zeta (1+\frac{1}{s}) - s$ is strictly decreasing. $\Box$

Theorem 3. Let $c_a$ be as defined in Lemma 2. The asymptotic expansion of $c_a$ is

$\displaystyle{c_a = 1-\gamma_0 + \frac{\gamma_1}{a-1} - \frac{\gamma_2 + \gamma_1-\gamma_1\gamma_0}{(a-1)^2}}$

$\displaystyle{+ \frac{\gamma_3 + 2\gamma_2- 2\gamma_2\gamma_0 + \gamma_1 - 2\gamma_1\gamma_0 + \gamma_1\gamma_0^2 - \gamma_1^2}{(a-1)^3}+ O\Big(\frac{1}{a^4}\Big)}.$

Proof. This is obtained by inverting the above Stieltjes series expansion of $\zeta (1+\frac{1}{a-1+c_a})$$\Box$

The next result gives us a method to compute $c_a$ numerically. We have

Lemma 4. Let $k_r$ be defined recursively as

$\displaystyle{k_{r+1} = a + k_r - \zeta \Big (\frac{a+k_r}{a-1+k_r}\Big )}$

where $k_0 = 1-\gamma_0$. Then $\lim_{r \rightarrow \infty}k_r = c_a$ .

Marek Wolf, has sent me the computed values of $c_n$ for $2 \le c_n \le 1000$. Some of these numbers were used in the proof of the following theorem.

Proof. Trivial. $\Box$

Theorem 5. If $1 \le p-q \le 40$ then $\zeta (\frac{p}{q}) \notin \mathbb N$.

Proof. We have $c_2 \approx 0.37240621590$. There are exactly twenty three positive fractions $\frac{m}{n}$ with $n \le 40$ such that

$\displaystyle{c_2 \le \frac{m}{n} < 1- \gamma_0}$.

For each of these fractions, we find (by actual computation) that there exists a integer $r \le 43$ such that

$\displaystyle{c_r \le \frac{m}{n} < c_{r+1}}$.

Since $c_n$ is strictly increasing, it implies that if $c_n \in \mathbb Q^+$ then the denominator of $c_n$ has to be $\ge 41$. Hence the theorem follows. $\Box$

If we continue working with the idea given the proof of Theorem 5, we can prove $p-q > n_0$ for some large natural number $n_o$. But this method gives no hope of proving that $\zeta (\frac{p}{q}) \notin \mathbb N$. Therefore we must look for some other ideas.

Theorem 6. Let $\zeta (1+\frac{1}{x_i}) \in \mathbb N$, $x_i \in \mathbb R^+ , i=1, 2, \ldots , k$ and $a_j \in \mathbb N$ such that

$\sum a_j = \{1, 2, 3, 4, 5, 7, 8, 10, 12 , 15\}$.

If $y=m+\sum a_k x_k$ where is any non-negative integer; then $\zeta (1+\frac{1}{y}) \notin \mathbb N$.

Some applications of Theorem 6 are given below.

Example 7. If $x \in \mathbb R^+$ and $2 \le n \le 5, n \in \mathbb N$ then at least one of the two numbers

$\zeta (1+\frac{1}{x})$ and $\zeta (1+\frac{1}{nx})$

is not a integer.

Example 8. If $x, y \in \mathbb R^+$ then at least one of the three numbers

$\zeta (1+\frac{1}{x}), \zeta (1+\frac{1}{y})$ and $\zeta (1+\frac{1}{x+y})$

is not a integer.

## 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)