## Sunday, 24 November 2013

### A little analysis problem

Recently I've been reading Springer GTM on hyperreals, but I'm not good enough to take any of these into the discussions here. I'm going to discuss a nice little problem here but due to the competitive nature of this question I am not going to disclose the source of it (maybe till the competition being due).

Question. Suppose $\left\{ a_n\right\} \subseteq \mathbb{N}$ strictly increasing. Show that $\sum_{i=1}^{\infty} \frac{a_{i+1}-a_i}{a_i}$ diverges.

A geometric interpretation gives you the conclusion immediately:

Solution. $\sum_{i=1}^{\infty} \frac{a_{i+1}-a_i}{a_i} \geq \sum_{i=1}^{\infty} \sum_{j=a_i}^{a_{i+1}-1} j^{-1} = \sum_{i = a_1}^{\infty} i^{-1}$

That clearly converges as indicated by the comparison test since the last one is a partial harmonic series. Please note that the above expression should be avoided as the $\geq$ sign is vague due to the divergence nature of the series.

Now...

Question 2. Suppose $\left\{ a_n\right\} \subseteq \mathbb{R}^+$ strictly increasing. Will $\sum_{i=1}^{\infty} \frac{a_{i+1}-a_i}{a_i}$ converge?

Extending the above proof to $\mathbb{R}$ seems to be a nice idea. To do this we shall compare the term $\frac{a_{i+1}-a_i}{a_i}$ with the integral $\int ^{a_{i+1}}_{a_i} x^{-1}dx$. We can show that they are usually asymptotically equivalent. Why usually? Let's bound them as the following:

The lower bound is clear (we use that for divergence). Suppose the sequence is unbounded then

$\sum_{i=1}^{\infty} \frac{a_{i+1}-a_i}{a_i} \geq \sum_{i=1}^{\infty} \int ^{a_{i+1}}_{a_i} x^{-1}dx = \int ^{\infty} _{a_1} x^{-1} dx$

Of course if it is bounded then it converges to $k\in \mathbb{R}$ then the series is lowerly bounded by $\int^k_{a_i} x^{-1} dx = \ln (k) - \ln (a_1)$.

What bothers us is the upper bound. Since $\left\{ a_n\right\}$ strictly increasing we use the convention that $k$ being limit of the sequence $\left\{ a_n\right\}$, which is infinity if it diverges.

Define $N_i = \frac{a_{i+1}}{a_i}$. Then if $\left\{ N_i\right\}$ upperly bounded by $N$ then the series is also upperly bounded by $\int ^{ k}_{a_1} Nx^{-1}dx = N(\ln k - \ln 1)$, so if $N_i$ and $a_i$ bounded then the series is bounded as well.

Is it possible that $a_i$ bounded but $N$ being unbounded? The answer is clear: no. Suppose $a_i$ upperly bounded by $A$. Then $N_i = \frac{a_{i+1}}{a_i} \leq \frac{A}{a_1}$ so that it is also upperly bounded.

Solution. The series converges if and only if the sequence $\left\{ a_n\right\}$ converges. i.e. iff the sequence is bounded above.

Define $f(x): \mathbb{R}^+ \rightarrow \mathbb{R}^+$ that $f(x) = \sum_{i: a_i \leq x} \frac{a_{i+1}-a_i}{a_i}$. Then it would be nice to do some asymptotic analysis over the function $f$ given that $a_i$ diverges.

From our previous analysis, if $N_i$ bounded then $f(x) \sim \log x$. What if $N$ unbounded? If we think about how $N_i$ is defined we can make up this example: $a_n = n!$ then $\frac{a_{i+1}-a_i}{a_i} = n$ .

Then it is clear that $N_i$ unbounded as $N_i = i+1$ with $f(n!)\sim n^2$, but if $N_i$ bounded we have $f(n!) \sim \log (n!) \sim n\log n$ by Stirling's formula, which has a different growth rate.

Foods for thought:

Given strictly increasing sequence $\left\{ a_n\right\}$ where $a_n \neq 0 \forall n\in \mathbb{N}$, define $b_i = \frac{a_{i+1}-a_i}{a_i}$.

1) Given a sequence $\left\{ a_n \right\} \in \mathbb{R}/\left\{ 0\right\}$. Give the condition that the series $\sum b_i$ converges.

2) Repeat Q1 if $\left\{ a_i\right\}\subseteq \mathbb{R}^+$ is increasing

3) For $k\in \mathbb{N}$, Define $b_{i,k} = \frac{a_{i+k} - a_i}{a_i}$ where $\left\{ a_i\right\}\subseteq \mathbb{R}^+$ strictly increasing. Determine the condition of convergence for $\sum_i b_{i,k}$.

4) Back to the original question. How fast can $f(x)$ grow? Are there any asymptotical boundings?