The Cauchy Proposition: Stare at It Until It Becomes Obvious!

柯西命题：盯着它到显然成立为止！ 苏剑林 · 2015-04-19 · Mathematics limit analysis trivially true

In the section on sequence limits in mathematical analysis, there is a very basic result known as the "Cauchy proposition":

If $\lim_{n\to\infty} x_n=a$, then
$$\lim_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n}=a$$

This post is about this proposition, and along with it, some similar problems.

Proof of the Cauchy Proposition

The proof of the Cauchy proposition isn't hard—it just follows directly from the definition of convergence. Since $\lim_{n\to\infty} x_n=a$, for any given $\varepsilon > 0$ there exists a sufficiently large $N$ such that for any $n > N$, we have

$$\left|x_n - a\right| < \varepsilon/2\quad(\forall n > N)$$

Then, for sufficiently large $n$, we have

$$\begin{aligned}&\left|\frac{x_1+x_2+\dots+x_n}{n}-a\right|\\ =&\left|\frac{(x_1-a)+(x_2-a)+\dots+(x_n-a)}{n}\right|\\ \leq &\left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right|\\ &\quad+\left|\frac{(x_{N+1}-a)}{n}\right|+\left|\frac{(x_{N+2}-a)}{n}\right|+\dots+\left|\frac{(x_{n}-a)}{n}\right|\\ < & \left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right|+\frac{n-N}{n}\frac{\varepsilon}{2}\\ < & \left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right|+\frac{\varepsilon}{2}\\ \end{aligned}$$

Taking $M > N >0$ large enough, as long as $n > M$, we have

$$\left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right| < \frac{\varepsilon}{2}$$

and hence

$$\left|\frac{x_1+x_2+\dots+x_n}{n}-a\right| < \varepsilon$$

Since $\varepsilon$ is arbitrary, it follows that

$$\lim_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n}=a$$

In fact, the intuitive meaning of the Cauchy proposition is also quite clear: it simply tells us that if a sequence converges, it becomes increasingly "flat," so that when we take averages, even if the sequence fluctuates wildly at the start, those fluctuations can eventually be smoothed out by taking a sufficiently large $n$.

A Variant

Now, let's have the reader try the following problem.

Given $\lim_{n\to\infty} x_n=a$, prove that
$$\lim_{n\to\infty}\frac{x_1+2 x_2+\dots+n x_n}{n^2}=\frac{1}{2}a$$

Please don't rush to pick up your pen—stare at it, stare at it hard, until the problem gets scared and hands you the answer itself!!

======== A Glorious Dividing Line ========

In fact, once we have the Cauchy proposition, this problem doesn't need to be proved at all—it's simply obvious! Why? Why should we bother considering the sequence $\{x_n\}$? Why not instead consider the following sequence?

$$\{y_n\}={x_1, x_2, x_2, x_3, x_3, x_3, x_4, x_4, x_4, x_4, x_5,\dots}$$

Clearly, we also have

$$\lim_{n\to\infty} y_n =a$$

and moreover

$$\frac{1}{n^2}\sum_{k=1}^n k a_k=\frac{1}{n^2}\sum_{k=1}^{\frac{1}{2}n(n+1)} y_n=\frac{\frac{1}{2}n(n+1)}{n^2}\frac{\sum_{k=1}^{\frac{1}{2}n(n+1)}y_n}{\frac{1}{2}n(n+1)}$$

By the Cauchy proposition, the limit of the summation term on the far right is immediately seen to be $a$, while the limit of $\frac{\frac{1}{2}n(n+1)}{n^2}$ is obviously $\frac{1}{2}$. Therefore

$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k a_k=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^{\frac{1}{2}n(n+1)} y_n=\frac{1}{2}a$$

So, whenever possible, try looking at things from a different angle—you might stumble upon something unexpected. Sometimes all it takes is staring at something long enough for it to become obvious. I think that being able to see the same problem from different angles is really what it means to fully understand it.