# If ${a_n}$ converges, then ${a_{2n}}$ converges proof question [duplicate]

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.

## marked as duplicate by user99914, JavaMan, heropup, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 7 '18 at 15:32
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, $|a_{2n}-L|<\varepsilon.$