[Euler-style Mathematics] Finding a Rigorous Answer

[欧拉数学]找出严谨的答案 苏剑林 · 2013-09-09 · Mathematics integral euler mathematics sequence

In a few earlier posts, I've talked about "Euler-style mathematics." Broadly speaking, Euler-style mathematics refers to creative, intuition-driven tricks and methods — approaches that can derive beautiful results even though the methods themselves aren't rigorous. However, in many cases, rigor and intuition are only a step apart. What follows is a problem from my Mathematical Analysis final exam last semester, and the inspiration for my solution came precisely from this kind of "Euler-style mathematics."

Let ${a_n}$ be an increasing sequence of positive numbers. Prove that $\sum\limits_{n=1}^{\infty}\left(1-\frac{a_n}{a_{n+1}}\right)$ converges if and only if ${a_n}$ converges.

I've heard that the official reference solution uses the Cauchy convergence criterion for sequences; I didn't look at it closely, since I was exploring a more intuitive approach of my own. This is exactly the spirit of "I do not understand what I cannot create." Here's how I thought about it.

Reflection

A rather powerful method for judging whether a series converges is the integral test, and that's usually the first thing that comes to mind for me. Although no explicit function is given here, so we can't actually integrate, we can still draw some inspiration from the idea.

Let $a_n \equiv a(n)$. When $a(n)$ varies slowly, the following approximations are quite good: $a'(x) \sim a(x+1)-a(x)$ and $a(x)\sim a(x+1)$, hence $1-\frac{a(x)}{a(x+1)}\sim \frac{a'(x)}{a(x)}$. By the integral test, we then have

$$S=\sum_{n=1}^{\infty}\left(1-\frac{a_n}{a_{n+1}}\right) \sim \int_1^{+\infty} \frac{a'(x)}{a(x)}dx = \ln a(x)\big|_1^{+\infty}$$

It's worth noting that every step here is non-rigorous, and might even fail to hold in general. But it lets us look at the problem from a different angle, and methodologically, each step is representative of a general technique — approximating a difference by a derivative, approximating a sum by an integral, and so on. The final result also faintly connects back to the problem itself: if $a(\infty)$ exists, then this integral exists, and so the series converges too.

Setting aside rigor for the moment, the most important thing this line of thought gives us is: $\ln a(x)$!

Sufficiency

This means precisely $S_n \sim \ln[a(n)]$. So we can consider

$$\ln a_{n+1} - \ln a_n = -\ln \frac{a_n}{a_{n+1}} = - \ln \left[1-\left(1-\frac{a_n}{a_{n+1}}\right)\right]$$

and we have $-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+...$, hence $-\ln(1-x) > x$, which lets us write

$$\ln a_{n+1} - \ln a_n > 1-\frac{a_n}{a_{n+1}}$$

Using this inequality, we obtain

$$\ln a_{n+1} - \ln a_1 >\sum_{i=1}^{n}(1-\frac{a_i}{a_{i+1}})$$

This shows that convergence of ${a_n}$ implies convergence of the series sum, which proves sufficiency.

(Before arriving at the result, we don't yet know whether we're going to end up proving sufficiency or necessity — that only becomes clear after the analysis. But either way, as long as we have an inequality like this, we can prove one direction or the other, and that's part of our confidence here. All that remains is to construct the reverse inequality.)

Necessity

To prove necessity, we need to find a similar inequality but with the sign reversed. Looking at $-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+...$, we see that no matter where we truncate it, we can't produce an inequality with the opposite sign. So instead consider $-\ln(1-x) < 2x$ — this doesn't always hold, but it does hold at least for $x \in\left[0,\frac{1}{2}\right]$. Moreover, given that $\sum\limits_{n=1}^{\infty}\left(1-\frac{a_n}{a_{n+1}}\right)$ converges, only finitely many terms of $1-\frac{a_n}{a_{n+1}}$ can be greater than $\frac{1}{2}$; otherwise we'd get a contradiction. So there exists some $N$ such that for $n > N$ we always have $1-\frac{a_n}{a_{n+1}} \in \left(0,\frac{1}{2}\right]$, which guarantees that $-\ln(1-x) < 2x$ applies. That is, for sufficiently large $n$, we have

$$\ln a_{n+1} - \ln a_n < 2\left(1-\frac{a_n}{a_{n+1}}\right)$$

From this we can then write

$$\ln a_{n+1} - \ln a_{N+1} < 2 \sum_{i=N+1}^{n}\left(1-\frac{a_i}{a_{i+1}}\right)$$

So convergence of the latter implies convergence of the former, which proves necessity.

Summary

Whatever the case, it's always beneficial to think in multiple directions. Being able to arrive at the same method as the reference solution is certainly nice, but we shouldn't confine ourselves to any one particular approach — instead, we should center our thinking on ourselves and find a framework of ideas that's genuinely our own. That's what it really means to create an answer. As for the "Euler-style mathematics" angle: no teacher is going to check what you scribbled on your scratch paper, so when working things out, brainstorm as freely as possible and try to associate as many methods as you can. As the saying goes, "a stone from another hill can be used to polish jade." It's worth noting that this process isn't confined to exams — it runs through our entire learning experience. And in that context, hindsight — figuring things out after the fact — turns out to be extremely valuable!