I wanted to prove the following:

if $x_n \to x$ then $y_n \to x$ where $$ y_n = {x_1 + \dots + x_n \over n}$$

Please can you tell me if my proof is correct? My proof is this:

Let $\varepsilon > 0$. Fix $N$ such that $n > N$ implies $|x_n - x| < {\varepsilon \over 2}$.

Then $\left |{1 \over n} \sum_{k=N+1}^{N + n} x_k - x \right | < {\varepsilon \over 2}$. Now let $M$ be such that ${|x_1 + \dots + x_N | \over M} < {\varepsilon \over 2}$ and $M > N$. Then $$ \left | \sum_{k=1}^M {x_k \over M} - x  \right | \le \left | \sum_{k=1}^N {x_k \over M} \right | + \left | \sum_{k=N+1}^M {x_k \over |M-N|} - x  \right | < \varepsilon $$ Here the proof is finished. But one can observe:

It is possible that $y_n$ converges even if $x_n$ doesn't: If $x_{2n} = 0$ and $x_{2n + 1} = 1$ then $x_n$ does not converge but $y_n \to {1 \over 2}$.


marked as duplicate by Guy Fsone, Sahiba Arora, Did real-analysis Jan 1 '18 at 22:47

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  • $\begingroup$ Why is $\left | \sum_{k=N+1}^M {x_k \over |M-N|} - x \right |<\frac{\varepsilon}{2}$? $\endgroup$ – sequence Sep 30 '16 at 9:49
  • $\begingroup$ @blue how do you obtain $$ \left | \sum_{k=1}^M {x_k \over M} - x \right | \le \left | \sum_{k=1}^N {x_k \over M} \right | + \left | \sum_{k=N+1}^M {x_k \over |M-N|} - x \right |?$$ $\endgroup$ – Saaqib Mahmood Dec 4 '16 at 7:18
  • $\begingroup$ See my proof here: math.stackexchange.com/questions/2440333/… $\endgroup$ – Guy Fsone Jan 1 '18 at 16:34

I think the proof is correct.

For your question, I think there are many way to sum up a divergent series like $-1+1-1-1\cdots$, which by some summation goes to $\frac{1}{2}$. The source you should consult as a reference is Stein's Fourier Analysis in his analysis book series. He discussed the convergence property of Caesaro means in detail. I do not have the book with me (I read it four years ago) and I do not recall the details. So that's all I can offer.

  • $\begingroup$ Thank you, also for a book recommendation!! $\endgroup$ – blue Nov 16 '13 at 21:26

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