Prove that $k_n$ $→L$ as $n→∞$. Let $(p_n)_n∈N$ be a sequence such that for all $n∈N$, $p_n ∈N$. We also assume that $p_n →+∞$ as $n→∞$. Let $(s_n)n∈N$ be a sequence of real numbers that converge to some L $∈ R$. Let $(k_n)$ $n∈N$ be the sequence defined by $k_n$ = $s_{p_n}$ for all $n∈N$. Prove that $k_n$ $→L$ as $n→∞$.
This is what I understand
$k_n$ is a subsequence of $s_n$ where the elements of the sequence $p_n$ is the index at which the elements of the sequence $s_n$ is chosen to construct the subsequence $k_n$. I am not sure if this is right but this is what I have based my attempt on. 
My attempt
To prove that $k_n$ converges to L as $n→∞$, we need to first prove that $n→∞$. To prove $n→∞$, the amount of indexes in $k_n$ is the same as the the elements of the sequence $p_n$. As the sequence $p_n$ diverges to $+∞$ and by the definition of the infinite limit, for any $M∈R$ there exists $N_M∈N$, for any n $\ge$ $N_M$ where $p_n$ $\ge$ M. Thus, this proves that there are infinite elements in the sequence $p_n$ as it is not bounded above and diverges to $+∞$.
By the subsequence limit theorem, if $s_n$ converges to L as $n→∞$ then the subsequence of $s_n$, such as $k_n$ must also converge to the same limit L. Thus proving that as $n→∞$ in $k_n$, then the limit of $k_n$ must be L. 
This is what I have so far and I am not sure if it looks right. I feel like my answer is too simple and more needs to be done. 
 A: The problem is in the understanding : $p_n$ could have repeated elements. On the other hand, a subsequence of a sequence does not repeat elements of the sequence or go back to earlier terms.
For example, let us take the sequence $s_n$ to be $s_n = \frac 1n$. Suppose I chose $p_n$ to be the sequence $1,1,2,1,2,3,1,2,3,4,...$ then the sequence $k_n$ will look like :
$$
1 , 1 , \frac 12, 1 , \frac 12 , \frac 13 , 1 , \frac 12 , \frac 13 , \frac 14, ...
$$
which is not a subsequence of $s_n$. 
The above $p_n$ does not satisfy the condition $p_n \to \infty$. So what does? Take $1,2,3,2,3,4,3,4,5,4,5,6,...$ : write down the sequence, it will consist of repeated elements, so won't be a subsequence. However, $p_n \to \infty$. If you write down the terms of $k_n$, it will also become apparent that $k_n \to 0$.
Thus, the last paragraph doesn't work, hence the proof fails.

Instead, keep it simple. Start with $\epsilon > 0$, you need to find $N$ so that whenever $n > N$ we have $|k_n - L| < \epsilon$.


*

*Since $s_n \to L$, we can find $N'$ so that $|s_n - L| < \epsilon$ for $n > N'$.

*Since $p_n \to \infty$, we can find $N$ so that $n > N \implies p_n > N'$.

*Show that if $n > N$ then $|k_n - L| < \epsilon$.

*Conclude.
