# 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.]

# Four Examples

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Following are the four examples of sequences (along with their properties) which can be helpful to gain a better understanding of theorems about sequences (real analysis):

• $\langle n\rangle_{n=1}^{\infty}$ : unbounded, strictly increasing, diverging
• $\langle \frac{1}{n}\rangle_{n=1}^{\infty}$ : bounded, strictly decreasing, converging
• $\langle \frac{n}{1+n}\rangle_{n=1}^{\infty}$ : bounded, strictly increasing, converging
• $\langle (-1)^{n+1}\rangle_{n=1}^{\infty}$ : bounded, not converging (oscillating)

I was really amazed to found that $x_n=\frac{n}{n+1}$ is a strictly increasing sequence, and in general, the function $f(x)=\frac{x}{1+x}$ defined for all positive real numbers is an increasing function bounded by 1:

The graph of x/(1+x) for x>0, plotted using SageMath 7.5.1

Also, just a passing remark, since $\log(x)< x+1$ for all $x>0$, and as seen in prime number theorem we get an unbounded increasing function $\frac{x}{\log(x)}$ for $x>1$

The plot of x/log(x) for x>2. The dashed line is y=x for the comparison of growth rate. Plotted using SageMath 7.5.1

# 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.).

# Kempner Series & MMC

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In Madhava Mathematics Competition 2015 (held in January 2015), we were asked to prove the convergence of Kempner Series (first time proved in 1914). Recently I discovered the paper by A. J. Kempner (http://www.jstor.org/stable/2972074), so in this blog post I will state and prove that problem.

The basic idea behind proof is to divide whole series into chunks (finding symmetry) and then construct a converging geometric series which will act as upper bound of Kempner Seires.

A detailed study of this cartoon has been done by Ben Orlin (http://bit.ly/1KD4shF)

Theorem (Kempner Series, 1914). Harmonic Series,$\sum_{k=1}^{\infty} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} +\ldots$, converge, if the denominators do not include all natural numbers $1,2,3,\ldots$, but only those numbers which do not contain any figure 9.

Proof: Given Series is:
$K = \frac{1}{1}+\ldots +\frac{1}{8}+\frac{1}{10}+\ldots +\frac{1}{18}+\frac{1}{20} + \ldots + \frac{1}{28}+\frac{1}{30}+\ldots +\frac{1}{38}+\frac{1}{40}+\ldots +\frac{1}{48}+\frac{1}{50} + \ldots + \frac{1}{58}+\frac{1}{60}+\ldots+\frac{1}{68}+\frac{1}{70}+\ldots +\frac{1}{78}+\frac{1}{80} + \ldots + \frac{1}{88}+\frac{1}{100}+\ldots$
Now we can rewrite above series as:
$S = a_1 + a_2 + \ldots + a_n + \ldots$
where $a_n$ is the sum of all terms in $K$ of denominator $d$ with $10^{n-1} \leq d < 10^{n}$.
Observe that, each term of $K$ which forms part of $a_n$, is less than or equal to $1/10^{n-1}$.

Now count the number of terms of $K$ which are contained in $a_1$, in $a_2$, $\ldots$, in $a_n$. Clearly, $a_1$, consists of 8 terms, and $a_1 < 8 \cdot 1 < 9$. In $a_2$ there are, as is easily seen, less than $9^2$ terms of $K$, and $a_2 < (9^2/10)$. Altogether there are in $K$ less than $9^2 + 9$ terms with denominators under 100.

Assume now that we know the number of terms in $K$ which are contained in $a_n$ to be less than $9^n$, for $n = 1, 2, 3, \ldots, n$. Then, because each term of $K$ which is contained in $a_n$ is not greater than $1/10^{n-1}$, we have $a_n < (9^n/10^{n-1})$, and the total number of terms in $K$ with denominators under $10^n$ is less than $9^n + 9^{n-1} + 9^{n-2} + \ldots + 9^2 + 9$.

Now, let’s go for induction. For $n = 1$ and $n = 2$ we have verified all this, and we will now show that if it is true for $n$, then $a_{n+1} < (9^{n+1}/10^n)$. $a_{n+1}$ contains all terms in $K$ of denominator $d$, $10^n \leq d < 1^{n+1}$. This interval for $d$ can be broken up into the nine intervals,  $\alpha \cdot 10^n \leq d < (\alpha + 1)10^n$, $\alpha = 1,2, \ldots, 9$. The last interval does not contribute any term to $K$, the eight remaining intervals contribute each the same number of terms to $K$, and this is the same as the number of terms contributed by the whole interval $0 < d < 10^n$, that is, by assumption, less than $9^n + 9^{n-1} + 9^{n-2} + \ldots + 9^2 + 9$.

Therefore, $a_{n+1}$ contains less than $8(9^n + 9^{n-1} + 9^{n-2} +\ldots + 9^2 + 9) < 9^{n+1}$ terms of $K$, and each of these terms is less than or equal to $1/10$, we have $a_{n+1} < (9^{n+1}/10^n)$.

Hence, $S = a_1 + a_2 + a_3 + \ldots < 9 + \frac{9}{10} + \ldots + \frac{9^{n+1}}{10^n}+ \ldots = 90$
Thus, $S$ converges, and since, $K=S$, $K$ also converges.

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Note: There is nothing special about 9 here, the above method of proof holds unchanged if, instead of 9, any other figure $1, 2, \ldots, 8$ is excluded, but not for the figure $0$.