Let $x_k \in \mathbb C$ for $k \in \mathbb N \cup {0}$ and let $y_k = \frac{(x_0 + x_1 + ... + x_k)}{k+1}$. We want to prove that if $x_k$ converges to $x$ ($x \in \mathbb C$) as $k \rightarrow \infty$ then $y_k$ also converges to $x$ as $k \rightarrow \infty$. This seems like a simple interesting exercise but having a little trouble. If $x_k \rightarrow x$ then from this I tried to make conclusions about convergence of $y_k$ since it depends of $x_k$. As $k \rightarrow \infty$ then $y_k \rightarrow \frac{(x_0 + x_1 + ... + x)}{k+1}$ but how can we conclude $\frac{(x_0 + x_1 + ... + x)}{k+1} = x$? Intuitively we also notice that all the finite terms over k+1 will go to 0 as $k \rightarrow \infty$.
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1Let me review duplicate, didn't see that before posting. – user77404 Jul 15 '13 at 00:01
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That sequence looks to be effectively the same. – dfeuer Jul 15 '13 at 00:06
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This question is a duplicate, I didn't notice the other answers, can I delete this post? – user77404 Jul 15 '13 at 00:22
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Don't delete the post yourself. If the moderators want, they can merge them to preserve the existing answer. – dfeuer Jul 15 '13 at 00:32
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We may start by proving the lemma that if $x_k \to 0$, then $y_k\to 0$.
Choose $N(\epsilon)$ such that $|x_k|<\epsilon$ for all $k\geq M$. Then let $M\in \mathbb{Z}^+$, and look at $y_{N+MN}=y_{N(1+M)}$. If we let $|y_N| = |A|$, then we have $$|y_{N+MN}|\leq \frac{A+MN\epsilon}{(M+1)N}.$$
Note that as $M\to\infty$, the first term goes to zero, and the second term goes to $\epsilon$.
The general case $x_k \to x$ can be shown by defining an auxiliary sequence $\mu_k := x-x_k$. Note $\mu_k$ goes to zero, and you can apply the lemma to $\mu_k$ to get the result you want, because $$y_k = L + (\text{averages of } \mu_k)$$
Eric Auld
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