# 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):

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

# Real Numbers

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Few days ago I found something very interesting on 9gag:

There are lots of interesting comments, but here is a proof from the comments:

…. Infinite x zero (as a limit) is indefinite. But infinite x zero (as a number) is zero. So lim( 0 x exp (x²) ) = 0 while lim ( f(X) x exp(X) ) with f(X)->0 is indefinite …

Though the statement made in the post is very vague and can lead to different opinions, like what about doing the product with surreal numbers, but we can safely avoid this by considering the product of real numbers only.

Now an immediate question should be (since every positive real number has a negative counterpart):

Is the sum of all real numbers zero?

In my opinion the answer should be “no”. As of now I don’t have a concrete proof but the intuition is:

Sum of a convergent series is the limit of partial sums, and for real numbers due to lack of starting point we can’t define a partial  sum. Hence we can’t compute the limit of this sum and the sum of series of real numbers doesn’t exist.

Moreover, since the sum of all “positive” real numbers is not a finite value (i.e. the series of positive real numbers is divergent) we conclude that we can’t rearrange the terms in series of “all” real numbers (Riemann Rearrangement Theorem). Thus the sum of real numbers can only be conditionally convergent. So, my above argument should work. Please let me know if you find a flaw in these reasonings.

Also I found following interesting answer on Quora:

The real numbers are uncountably infinite, and the standard notions of summation are only defined for countably many terms.

Note: Since we are dealing with infinite product and sum, we can’t argue using algebra of real numbers (like commutativity etc.).