I'm trying to solve an Analysis problem.

I have a given sequence $(x_n)_n$ and $(y_n)_n$ defined by $ y_n = 1/n \sum _{ k=1 }^{ n }{ x_{ n } } $. I have to prove

  1. ${ ({ x }_{ n }) }_{ n }converges\quad \Longrightarrow { ({ y }_{ n }) }_{ n }converges$

I haven't found any counterexamples for it so I tried to prove it but i don't know how to start. Any ideas?

  • $\begingroup$ Just for clarification, you mean that $y_n$ is the arithmetic mean of $x_1, \ldots, x_n$, right? That is, $$y_n = \frac{1}{n} \Sigma_{k=1}^n x_k$$ right? $\endgroup$ – TM Gallagher Mar 6 at 3:14

If $x_n$ converges to $a$, then for any $\varepsilon$ there is an $N$ such that for all $n>N$, $|x_n-a|<\varepsilon$. Thus if the sum of the first $N$ terms is $s$, then $$\frac{s+k(a-\varepsilon)}{N+k}\leq y_{N+k}\leq \frac{s+k(a+\varepsilon)}{N+k}$$

Then the $s/(n+k)$ term approaches 0, and $k/(N+k)$ approaches 1, thus for any $\varepsilon$, there is some $M$ such that $a-2\varepsilon<y_M<a+2\varepsilon$, so the limit $b$ of $y_n$ equals a.

However, the converse is not true: consider $x_n=(-1)^n$. Hope this helps!


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