Let $\{a_n\}_{n \in \Bbb N}$ be convergent. Prove/Disprove: $\lim\limits_{n\to\infty}$ $|a_n - a_{2n}| = 0$ My main source of confusion is the use of $a_{2n}$. I’m not sure if this refers to the even entries or something else.
I know that since $a_n$ converges then $d(a_n,a) < \varepsilon$, or $\lim\limits_{n\to\infty}$ $a_n = a$.
If $a_{2n}$ refers to the subsequence of even entries it is straightforward to show that a sequence only converges if all of its subsequences converge.
 A: Yes $a_{2n}$ refers to the even entries of the sequence $\{a_n \in \Bbb R\}_{n \in \Bbb N}$.
And you are essentially begin asked to prove that the sequence $b_n = |a_n - a_{2n}|$ derived from our original one converges to $0$. And this can be accomplished by a standard $\varepsilon$-$N$ argument:
Well fix any $\varepsilon > 0$. If $a_n$ is convergent (to $a \in \Bbb R$ say), then by definition of convergence there is a natural $N$ s.t. $$|a_n - a| < \frac{\varepsilon}{2}$$ for all naturals $n > N$.
But look: for any natural $n$ obviously $2n \geq n$. Hence if $n > N$, then certainly $2n > N$ also. Thus for that same $\varepsilon$, you also get $$|a_{2n} - a| < \frac{\varepsilon}{2}$$ as long as $n > N$ remains true.
So now by the triangle inequality,
\begin{align*}
|b_n - 0| = |a_n - a_{2n}| &\leq |a_n - a| + |a - a_{2n}| \\
&= |a_n - a| + |a_{2n} - a| \\
&< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
\end{align*} for all naturals $n > N$. But that is exactly what $\lim\limits_{n \to \infty} b_n = 0$ means.
A: By definition, $(a_{2n}) = (a_2, a_4, a_6, a_8, \dots).$ 
Note the following: If $(a_n)$ converges, say to $a$,  then every subsequence of $(a_n)$ also converges to $a$, so $\lim_{n \to \infty} (a_n - a_{2n}) = a - a = 0$. Then one has the following standard theorem: If $(a_n)$ converges to $a$, the sequence $(\vert a_n \vert)$ converges to $\lvert a \rvert$. It follows
$$\lim_{n \to \infty} \lvert a_n - a_{2n} \rvert = \lvert a - a \rvert = \lvert 0 \rvert = 0.$$
A: The limit of $a_{2n}$ is the same as the limit of $a_n$, given the convergence of the sequence.
Thus you can apply standard theorems on limits to prove the statement. 
A: Let $\varepsilon>0$ be given. We want to prove that $\exists N$ so that $\forall n > N$ we have that
$$||a_{n}-a_{2n}|-0|=|a_n-a_{2n}|<\varepsilon$$
But the convergence of $(a_n)$ means that it's Cauchy, i.e. we can find an $N_0$ for our $\varepsilon$ so that $\forall n,m>N_0$ we have that $|a_{n}-a_{m}|<\varepsilon$. So let $N=N_0$, and let $m=2n$, and we are done.
A: One more answer:
1) $a_n$ converges to L.
2) Then the subsequence $a_{2n}$ converges to the same limit L.
3) $b_n:=a_n - a_{2n}$;
4) $\lim_{n \rightarrow \infty}  b_n= \lim_{n \rightarrow \infty}(a_n-a_{2n})$
$=\lim_{n \rightarrow \infty}a_n -\lim_{n \rightarrow \infty}a_{2n}=$
$L-L=0.$
5) $\lim_{n \rightarrow \infty}b_n=0$:
Given an $\epsilon >0$ there is a $n_0$ s t. for $n \ge n_0$
$|a_n-a_{2n}| = |b_n| < \epsilon$.
