joseph-nebus | Gaurish4Math

About 2.5 years ago I had promised Joseph Nebus that I will write about the interplay between Bernoulli numbers and Riemann zeta function. In this post I will discuss a problem about finite harmonic sums which will illustrate the interplay.

Consider the Problem 1.37 from The Math Problems Notebook:

Let $\{a_1, a_2, \ldots, a_n\}$ be a set of natural numbers such that $\text{gcd}(a_i,a_j)=1$ , and $a_i$ are not prime numbers. Show that $\displaystyle{\frac{1}{a_1}+\frac{1}{a_2}+ \ldots + \frac{1}{a_n} < 2}$

Since each $a_i$ is a composite number, we have $a_i = p_i q_i s_i$ for some, not necessarily distinct, primes $p_i$ and $q_i$ . Next, $\text{gcd}(a_i,a_j)=1$ implies that $p_i,q_i \neq p_j, q_j$ . Therefore we have:

$\displaystyle{\sum_{i=1}^n \frac{1}{a_i} \leq \sum_{i=1}^n \frac{1}{p_i q_i} \leq \sum_{i=1}^n \frac{1}{(\text{min}(p_i,q_i))^2} < \sum_{k=1}^\infty \frac{1}{k^2}}$

Though it’s easy to show that $\sum_{k=1}^\infty \frac{1}{k^2} < \infty$ , we desire to find the exact value of this sum. This is where it’s convinient to recognize that $\boxed{\sum_{k=1}^\infty \frac{1}{k^2} = \zeta(2)}$ . Since we know what are Bernoulli numbers, we can use the following formula for Riemann zeta-function:

$\displaystyle{\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}}$

There are many ways of proving this formula, but none of them is elementary.

Recall that $B_2 = 1/6$ , so for $n=1$ we have $\zeta(2) = \pi^2/6\approx 1.6 < 2$ . Hence completing the proof

$\displaystyle{\sum_{i=1}^n \frac{1}{a_i} <\zeta(2) < 2}$

Remark: One can directly caculate the value of $\sum_{k=1}^\infty \frac{1}{k^2}$ as done by Euler while solving the Basel problem (though at that time the notion of convergence itself was not well defined):

M	T	W	T	F	S	S
		1	2	3	4	5
6	7	8	9	10	11	12
13	14	15	16	17	18	19
20	21	22	23	24	25	26
27	28	29	30	31

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