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I am trying to prove that if a sequence $\{a_n\}$ converges, then the sequence $\{a_{2n}\}$ converges as well using the definition of convergence.

What I have so far is if $\{a_n\}$ converges, say to $L$, then for any given $\epsilon >0$, there exists an $n^* \in \mathbb{N}$ such that if $n > n^*$, then $\mid(a_n - L)\mid < \epsilon$. I think I want to choose an $n_1^*$ such that this is true for $\{a_{2n}\}$. I'm just not sure exactly how to choose the $n_1^*$. I thought maybe $\frac{n^*}{2}$, but I am not sure this is right and wouldn't know how to implement it.

Thanks in advance!

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marked as duplicate by user99914, JavaMan, heropup, José Carlos Santos real-analysis Sep 7 '18 at 15:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See, in particular, the answer here $\endgroup$ – user99914 Sep 6 '18 at 2:42
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Let $n^*$ be so that for any $n\geq n^*$ then $|a_n-L|<\varepsilon.$ Using this same $n^*$ it follows that $2n>n \geq n^*,$ and consequently, $|a_{2n}-L|<\varepsilon.$

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