$(a_{(k_n)})_{n\in\mathbb N}$ converges to L?

Question

The problem: Suppose $(a_n)_{n\in\mathbb N}$ is a sequence in $\mathbb R$ moreover that $(a_n)_{n\in\mathbb N}$ converges to $L \in \mathbb R$. $(k_n)_{n\in\mathbb N}$ is a sequence in $\mathbb N$, with $\lim_{n \to \infty}\frac{1}{k_n} = 0$. How can we show $(a_{(k_n)})_{n\in\mathbb N}$ converges to L?

My progress: Intuitively I can see why this is true moreover, after seeing a series of classes full of epsilon delta proofs of different limit theorems within sequences in general. I would bet that this will be a proof with creative uses of the epsilon delta definitions perhaps more than once in a nested fashion.

I've spent about 7-8hrs today on this question and have literally made no progress other than trying to bend the definition of convergence around like a mad man.

Answer 1

Let $\epsilon >0$. There exists $m$ such that $|a_n-L| <\epsilon$ for $n >m$. There exists $p$ such that $\frac 1 {k_n} < \frac 1 m$ for $n>p$. If $n >p$ then $|a_{k_n} -L| < \epsilon$.

Answer 2

If $\lim_{n} \frac{1}{k_n} \to 0$, then it is obvious that $\lim_{n} k_n \to \infty$ (you can use simple $\epsilon-\delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $\{a_{k_n}\}$ is just a subsequence, so the result follows.