I'm studying Real Analysis, and one problem gives me a trouble. The problem is as below:

Let $\{x_n\}$ be a sequence defined on $\mathbb{R}$ with $\displaystyle \lim_{n \to \infty} x_n = x$ for some $x \in \mathbb{R}$. Define a sequence $\{\sigma_n \}$ on $\mathbb{R}$ by $$\sigma_n= \frac{1}{n} ( x_1 + x_2 + x_3 + \cdots + x_n)$$ Find the flaw of the proof below, which tries to show the claim.

Claim : The sequence $\{\sigma_n\}$ converges. In addition, $\displaystyle \lim_{n \to \infty} \sigma_n = x$.

Proof. Since $\displaystyle \lim_{n \to \infty} x_n = x$, for any $\epsilon >0$, there exists a natural number $N$ such that $$n >N \quad \Rightarrow \quad \lvert x_n-x \rvert < \epsilon$$ Now fix $\epsilon >0$, and let $N_\epsilon$ be the natural number that satisfies the property above. Note that $$\lvert \sigma_n -x \rvert = \lvert\frac{1}{n} (x_1 + x_2 + \cdots + x_n) - x\rvert \leq \frac{1}{n} (\lvert{x_1-x}\rvert + \cdots + \lvert{x_n-x}\rvert)$$ Now, for sufficiently large $n>N_\epsilon$, we can divide the term above as $$\lvert \sigma_n -x \rvert = \frac{1}{n}(\lvert{x_1-x}\rvert + \lvert{x_2 - x}\rvert + \cdots + \lvert{x_{N_\epsilon}-x}\rvert)+\frac{1}{n}(\lvert{x_{N_\epsilon+1}-x}\rvert + \cdots + \lvert{x_n-x}\rvert)$$ Since the first term above has only finite constant terms, $$\frac{1}{n}(\lvert{x_1-x}\rvert + \lvert{x_2 - x}\rvert + \cdots + \lvert{x_{N_\epsilon}-x}\rvert) \to 0 \quad \text{as} \quad n \to \infty$$ Now, $$\lvert \sigma_n -x \rvert = \frac{1}{n}(\lvert{x_{N_\epsilon+1}-x}\rvert + \cdots + \lvert{x_n-x}\rvert) < \frac{1}{n} \times \epsilon (n-N_\epsilon) \to \epsilon$$ as $n \to \infty$. Therefore $\displaystyle \lim_{n \to \infty} \sigma_n = x$.

I understand that there is some problem in the proof, but I cannot clearly explain the answer! I think the problem comes from finding the limit not at once, but calculating the parts first. Could somebody explain this to me plainly?

  • 1
    $\begingroup$ $s=x{{{{{}}}}}$? $\endgroup$ Jun 16, 2019 at 17:07
  • $\begingroup$ I don't know if this is the "error" you are looking for but you say that the limit of the second sequence is "s" but have not defined "s"! The limit of the sequence should be "x" as Lord Shark the Unknown suggested. $\endgroup$
    – user247327
    Jun 16, 2019 at 17:12
  • $\begingroup$ Oh, that’s my typo, sorry. The limit of given should also be x! $\endgroup$
    – Alex Lee
    Jun 16, 2019 at 17:14
  • $\begingroup$ Would you mind fixing the parenthesis on the first $|\sigma_n - x| =$ line ? $\endgroup$
    – DanielV
    Jun 16, 2019 at 17:14
  • $\begingroup$ There are so many typos in there! I also edited it. Thanks for the comment. $\endgroup$
    – Alex Lee
    Jun 16, 2019 at 17:28

1 Answer 1


The mistake is in the line

Now, $$\lvert \sigma_n -x \rvert = \frac{1}{n}(\lvert{x_{N_\epsilon+1}-x}\rvert + \cdots + \lvert{x_n-x}\rvert) < \frac{1}{n} \times \epsilon (n-N_\epsilon) \to \epsilon$$ as $n \to \infty$.

because you lost the initial $\frac{1}{n}(\lvert{x_1-x}\rvert + \lvert{x_2 - x}\rvert + \cdots + \lvert{x_{N_\epsilon}-x}\rvert)\neq 0$ in $$\lvert \sigma_n -x \rvert = \frac{1}{n}(\lvert{x_1-x}\rvert + \lvert{x_2 - x}\rvert + \cdots + \lvert{x_{N_\epsilon}-x}\rvert)+\frac{1}{n}(\lvert{x_{N_\epsilon+1}-x}\rvert + \cdots + \lvert{x_n-x}\rvert)$$

It is easy enough to fix this though.

  • $\begingroup$ Thanks for the comment. I have additional question, if the proof from the line you quoted is revised as taking n→∞ directly to the expression |σn−x|=... and saying that the absolute value converges to zero since the first finite terms are constants divided by n and others are less than ϵ/n, is there still some flaw or gaps in the proof? $\endgroup$
    – Alex Lee
    Jun 16, 2019 at 17:27
  • $\begingroup$ You mean $\lvert\sigma_n-x\rvert\leq \frac{C(\varepsilon)}{n}+\varepsilon(1-\frac{N}n)$? That is fine. $\endgroup$ Jun 16, 2019 at 17:45

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