By 苏剑林 | April 19, 2015
In the section on limits of sequences in mathematical analysis, there is a very fundamental "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 article specifically discusses this proposition, as well as some similar problems.
Proof of the Cauchy Proposition
The proof of the Cauchy proposition is not difficult. We only need to use the definition of limit 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
\[|x_n - a| < \varepsilon/2 \quad (\forall n > N)\]
Then, for a sufficiently large $n$, we have
\[\begin{aligned}
&\left|\frac{x_1+x_2+\dots+x_n}{n}-a\right| \\
=&\left|\frac{(x_1-a)+(x_2-a)+\dots+(x_n-a)}{n}\right| \\
\leq &\left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right| \\
&\quad+\left|\frac{(x_{N+1}-a)}{n}\right|+\left|\frac{(x_{N+2}-a)}{n}\right|+\dots+\left|\frac{(x_{n}-a)}{n}\right| \\
< & \left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right|+\frac{n-N}{n}\frac{\varepsilon}{2} \\
< & \left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right|+\frac{\varepsilon}{2}
\end{aligned}\]
Choose a sufficiently large $M > N > 0$ such that for any $n > M$, we have
\[\left|\frac{(x_1-a)+(x_2-a)+\dots+(x_N-a)}{n}\right| < \frac{\varepsilon}{2}\]
Thus
\[\left|\frac{x_1+x_2+\dots+x_n}{n}-a\right| < \varepsilon\]
By the arbitrariness of $\varepsilon$, we know
\[\lim_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n}=a\]
In fact, the intuitive meaning of the Cauchy proposition is quite clear. It simply tells us that if a sequence is convergent, it becomes increasingly "flat." If we take the average, even if there are large fluctuations at the beginning, they can be smoothed out by a sufficiently large $n$.
A Variation
Below, I invite the reader to try this problem.
Given $\lim_{n\to\infty} x_n=a$, prove that
\[\lim_{n\to\infty}\frac{x_1+2 x_2+\dots+n x_n}{n^2}=\frac{1}{2}a\]
Reader, please do not be in such a hurry to pick up your pen. Stare at it, stare at it intently, until the problem is afraid of you and hands over the answer itself!!
======== Gorgeous Divider ========
In fact, with the Cauchy proposition, this problem does not need to be "proven" at all; it is obviously true! Why? Why should we bother considering the sequence $\{x_n\}$? Why not consider the following sequence instead?
\[\{y_n\}=\{x_1, x_2, x_2, x_3, x_3, x_3, x_4, x_4, x_4, x_4, x_5, \dots\}\]
Clearly, we also have
\[\lim_{n\to\infty} y_n = a\]
And furthermore
\[\frac{1}{n^2}\sum_{k=1}^n k x_k = \frac{1}{n^2}\sum_{k=1}^{\frac{1}{2}n(n+1)} y_k = \frac{\frac{1}{2}n(n+1)}{n^2} \frac{\sum_{k=1}^{\frac{1}{2}n(n+1)} y_k}{\frac{1}{2}n(n+1)}\]
From the Cauchy proposition, we immediately know that the limit of the summation term on the far right is $a$, and the limit of $\frac{\frac{1}{2}n(n+1)}{n^2}$ is obviously $\frac{1}{2}$. Therefore
\[\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k x_k = \lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^{\frac{1}{2}n(n+1)} y_k = \frac{1}{2}a\]
So, looking at things from a different perspective can lead to unexpected insights. Sometimes, if you just keep staring at it, it becomes obvious. I believe that being able to understand the same problem from different angles is what it means to truly see through a problem.