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This was a homework question in a first course in real analysis that I had taken as an undergraduate. The question was exercise 2.3.11 from Stephen Abbott's "Understanding Analysis" 2nd edition. It stuck with me because it was asked in the section immediately before he introduce infinite series, and so the author expects the student to only use the tools up to that point in the book without infinite series. (and at the time it was one of the only questions I couldn't finish myself.)

It was stated as follows:

Prove If $(x_n)$ is a convergent sequence, then the sequence $y_n = \frac{x_1 + \ \cdots \ + x_n}{n}$ converges to the same limit.

So my question is how do we prove this when limiting ourselves to only ideas in the section and those preceding that the problem is contained in?

So only using the definition of convergence of a sequence, uniqueness of limits, every convergent sequence is bounded, the algebraic limit theorem, and the order limit theorem (or anything else from page 1 to 55, including exercises) But WITHOUT using the monotone convergence theorem, the Bolzano-Weierstrass theorem, Cauchy sequences, or infinite series (or anything else past page 55.)

My solution: (I think this is right, check my work!)

Let $\varepsilon >0$.

Since $(x_n)$ converges, there exists a real number $x$ that $(x_n)$ converges to and a natural number $N$ such that for all $k \ge N$

$|x_k - x| < \frac{\varepsilon}{4}$. Fix this $N$ for the given $\varepsilon$.

Let $M = max\{|x_1-x|,\dots,|x_{N-1}-x|\}$.

Now choose a natural number $N_0 > \frac{8(N-1)M-2N\varepsilon}{4\varepsilon}$.

Let $n$ be a natural number such that $n \ge N_0$, then $n > \frac{8(N-1)M-2N\varepsilon}{4\varepsilon}$ which gives (by dividing both sides by $n$ and $\frac{\varepsilon}{2})$

$\frac{4(N-1)M-N\varepsilon}{4n} < \frac{\varepsilon}{2}$, and then add $\frac{\varepsilon}{2}$

$\frac{4(N-1)M-N\varepsilon}{4n} + \frac{\varepsilon}{2} < \varepsilon$.

Then notice $\frac{\varepsilon}{4n} + \frac{n\varepsilon}{4n} < \frac{\varepsilon}{2}$ so that we get

$|y_n - x| = |\frac{x_1 + \ \cdots \ + x_n}{n} -x| \le |\frac{x_1-x}{n}| + \cdots +|\frac{x_{N-1}-x}{n}| + |\frac{x_N-x}{n}| + \cdots + |\frac{x_n-x}{n}| < \frac{M}{n} + \cdots + \frac{M}{n} + \frac{\varepsilon}{4n} + \cdots + \frac{\varepsilon}{4n}$

$= \frac{M(N-1)}{n} + \frac{(n-N+1)\varepsilon}{4n} = \frac{M(N-1)}{n} - \frac{N\varepsilon}{4n} + \frac{\varepsilon}{4n} + \frac{n\varepsilon}{4n} < \frac{4(N-1)M-N\varepsilon}{4n} + \frac{\varepsilon}{2} < \varepsilon$

Therefore $(y_n)$ converges to $x$.

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    $\begingroup$ Sheldon Abbott? $\endgroup$
    – Snoop
    Mar 18, 2023 at 23:34
  • $\begingroup$ oops, my memory failed me $\endgroup$
    – mebenot
    Mar 18, 2023 at 23:55

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Yes, your proof is correct (except for the fact that your $\frac{\varepsilon}{4}$ in the second-to-last line should actually be $\frac{\varepsilon}{4n}$), and the idea is the right one - however, it can be made a bit more concise:

Let $\varepsilon>0$ be given, and $N\in\mathbb{N}$, so that $|x_n-x|\leq\frac{\varepsilon}{2}$ for all $n\geq N$. Also, let $M:= \max\{|x_1-x|,...,|x_{N}-x|\}$. Then for all $n\in\mathbb{N}$ with $n\geq N_0 := \frac{2NM}{\varepsilon}$:

$$|y_n -x| \leq \frac{\overbrace{|x_1-x|}^{\leq M}+...+\overbrace{|x_{N}-x|}^{\leq M}}{n} + \frac{|x_{N+1}-x|+...+|x_n-x|}{n}\leq \frac{NM}{n} + \frac{n\varepsilon}{2n} \leq \varepsilon$$ for all

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    $\begingroup$ Ah thanks, I fixed my error. You're right it's a bit neater to split the tail of the sum after $N$ instead before it. $\endgroup$
    – mebenot
    Mar 19, 2023 at 0:09

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