convergence – Some Mathematical Farts

Just to clarify all the numbers in this post are going to exist in $\mathbb{R}$ . Given a sequence $\{a_n\}$ where $n \in \mathbb{N}$ we have a collection of partial sums $\{s_n\}$ indexed by $\mathbb{N}$ and defined by $s_n = a_1 + a_2 + ... + a_n$ . If the sequence $\{s_n\}$ converges to $s$ (In the usual $\epsilon$ – $\delta$ way) we say the series converges and write $\sum_{n = 1}^{\infty} a_n =s$ . For completness if the sequence $\{s_n\}$ diverges, the series is said to diverge.

If you know sequences really well, you know series really well as every theorem about sequences can be stated in terms of series (putting $a_1 = s_1$ and $a_n = s_n - s_{n-1}$ for $n > 1$ ). In particular the monotone convergence theorem has an instantaneous counterpart for series.

Theorem: A series of non-negative real numbered terms converges if and only if the partial sums form a bounded sequence.

I’m going to omit the proof here but it is a quick application of the monotone convergence theorem to the partial sums. So why bring this up? Well if we impose that the terms in our series are decreasing monotically (which can appear in applications) we can apply the following theorem of Cauchy. What is interesting about this theorem is that a ‘thin’ subsequence of $\{a_n\}$ determines the convergence or divergence of the series.

Theorem: Suppose $a_1 \geq a_2 \geq ... \geq 0$ are real numbers. Then the series $\sum_{n=1}^{\infty} a_n$ converges if and only if the series $\sum_{k=0}2^k a_{2^k}$ converges.

Proof: By the previous theorem it suffices to consider only the boundedness of the partial sums. Let us write $s_n = a_1 + ... + a_n$ and $t_k = 2a_2 + ... + 2^ka_{2^k}$ . We will look at two cases, when $n < 2^k$ and when $n > 2^k$ .

For $n < 2^k$ we have $s_n \leq a_1 + (a_2 + a+3) + ... + (a_{2^k} + ... +a_{2^{k+1} - 1}) \leq a_1 + 2a_2 + ... + 2^ka_{2^k} = t_k$ where the first inequality followed from $n < 2^k$ and the second inequality from the hypothesis.

When $n > 2^k$ we have $s_n \geq a_1 + a_2 + (a_3 + a_4) + ... +(a_{2^{k-1} +1} + ... a_{2^k}) \geq \frac{1}{2}a_1 + a_2 +2a_4 + ... + 2^{k-1}a_{2^k}$ $= \frac{1}{2}t_k$ where the first inequality follows from $n > 2^k$ and the second (you guessed it) follows from our hypothesis.

Bringing these together we conclude that the sequence $\{s_n\}$ and $\{t_k\}$ are either BOTH bounded or BOTH unbounded which completes the proof.

When I came across this I thought it was pretty astounding (hence why it has made it onto the blog) so lets see it it action. We will use it to deduce for $p \in \mathbb{Z}$ that $\sum_{n=2}^{\infty} \frac{1}{n(log \:n)^p}$ converges if $p >1$ and diverges if $p \leq 1$ .

The monotonicity of the logarithmic function implies $\{\mbox{log } n \}$ increases which puts us in good position to apply our theorem. This leads us to the following which is enough as a proof.

$\sum_{k=1}^{\infty}2^k\frac{1}{2^k(log \:2^k)^p} = \sum_{k=1}^{\infty} \frac{1}{k(log \:2)^p} =\frac{1}{(log \; 2)^p} \sum_{k=1}^{\infty} \frac{1}{k^p}$ .

Suppose one day you get bored of the triangle inequality. You have done so much real analysis that you cannot see anything new and interesting in $|a-b| \le |a-z| + |z+b|$ anymore. But you want to keep doing analysis.

What can we do about this? One solution may be making this inequality more strict, and see where it takes us. For example, you may consider the following: $|a-b| \le \max{\{|a-z|,|z-b|\}}$ . Note that this inequality does not hold under our usual absolute value, for example: $5 = |8 - 3| \not \le \max{ \{ |8-4|, |4-3| \} } = \max{ \{4,1\} } = 4$ . But does this tweaking of the triangle inequality add anything new? It may not look like it at first, but it is a dream come true: a sequence is Cauchy if and only if distance of consecutive terms goes to zero! More mathematically: the sequence $(a_n)_{n \in \mathbb{N}}$ is Cauchy $\iff$ $|a_n-a_{n+1}| \to 0$ as $n \to \infty$

This is not true in the case of our usual absolute value. For example, the infamous sequence $H_n = \sum_{k = 1}^{n} \frac{1}{k}$ is not Cauchy in the reals $\mathbb{R}$ , for if it were Cauchy, it would converge since the reals numbers are complete. However, the difference between consecutive terms is $|1/n|$ which clearly tends to zero.

Before we dive into the proof of the previous claim, we need to note something: we are talking about convergence in $\mathbb{Q}$ however this would work in any metric space $(X,d)$ where the triangle inequality satisfies $d(a,b) \le \max{\{d(a,z), d(z,b) \}}$ for any $a,b,z \in X$ . This is because an absolute value on $\mathbb{Q}$ induces a metric by setting $d(a,b) = |a-b|$ .

Theorem: Let $| \cdot |$ be an absolute value on $\mathbb{Q}$ satisfying $|a+b| \le \max{ \{ |a|,|b|\} }$ . Then any sequence $(a_n)_{n \in \mathbb{N}}$ of rational numbers is Cauchy if and only if $|a_n-a_{n+1}| \to 0$ as $n \to \infty$ .

Proof: $( \implies)$ Let $\epsilon > 0$ . Since $(a_n)_{n \in \mathbb{N}}$ is Cauchy there exists $N \in \mathbb{N}$ such that for all $n,m > N$ we have $|a_n - a_m| < \epsilon$ . In particular, whenever $m = n+1$ , we obtain $|a_n - a_{n+1}| < \epsilon$ , which means that $|a_n - a_{n+1}| \to 0$ as $n \to \infty$ .

$( \impliedby )$ Let $\epsilon > 0$ . There exists $N \in \mathbb{N}$ such that $|a_n - a_{n+1}| < \epsilon$ for any $n > N$ .
Then, for all $m \ge n > N$ , applying our strict triangle inequality:

$\begin{aligned} |a_n - a_m| &= |a_n - a_{n+1} + a_{n+1} - \dots + a_{m-1} -a_{m}| \\ &\le \max{\{|a_n - a_{n+1}|, \dots, |a_{m-1} -a_{m}| \} } < \epsilon \end{aligned}$ .

Therefore the sequence is Cauchy. $\square$

This is all well and good, however, it makes us wonder whether there is any absolute value on $\mathbb{Q}$ with this strict triangle inequality property. The answer is yes, in fact, there are infinitely many of them, but this is the topic for another post. Until then, to satisfy your curiosity I will write down one such absolute value and I will leave it up to you to check that it is indeed an absolute value:

$|x|_3 = \frac{1}{3^{v(x)}}$
where $v(x) = n$ is the exponent of $3$ whenever we write the rational number as $x = 3^n \frac{p}{q}$ , where $p,q$ are coprime to $3$ .

Can you suggest a different absolute value?

Tag: convergence

A theorem by Cauchy on ‘thinning’ sequences

Unusual metrics on the rationals