Tag Archives: finite sum

Dec20

Riemann zeta function

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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):

pjimage

The Pleasures of Pi, E and Other Interesting Numbers by Y E O Adrian [Copyright © 2006 by World Scientific Publishing Co. Pte. Ltd.]

Jun25

Finite Sum & Divisibility

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I wish to discuss a small problem from The USSR Olympiad Problem Book (problem 59) about the finite sum of harmonic series. The problem asks us to prove that

\displaystyle{\sum_{k=2}^{n} \frac{1}{k}}  can never be an  integer for any value of n.

I myself couldn’t think much about how to prove such a statement. So by reading the solution, I realised that how a simple observation about parity leads to this conclusion.

Firstly, observe that among the natural numbers from 2 to n there is exactly one natural number which has the highest power of 2 as its divisor. Now, while summing up the reciprocals of these natural numbers we will get a fraction as the answer. In that fraction, the denominator will be an even number since it’s the least common multiple of all numbers from 2 to n. And the numerator will be an odd number since it’s the sum of (n-2) even numbers with one odd number (corresponding to the reciprocal of the number with the highest power of 2 as the factor). Since under no circumstances an even number can completely divide an odd number, denominator can’t be a factor of the numerator. Hence the fraction can’t be reduced to an integer and the sum can never be an integer.