Show that if ${s_{n_k}}_k$ converges to $L$, then ${s_n}_n$ converges to $L$.

Suppose a sequence ${s_n}_n$ is monotone increasing, consider any of its subsequences ${s_{n_k}}_k$. Show that if ${s_{n_k}}_k$ converges to $L$, then ${s_n}_n$ converges to $L$.

So I tried this problem and I was wondering if I did it correctly. My problem is that I’m unsure about the part where I say $s_n leq s_{n_{k’}}$. Any corrections would be appreciated, thanks!

$|s_n – L| = |s_n – s_{n_k} + s_{n_k} – L| leq |s_n – s_{n_k}| + |s_{n_k} – L|$

First note that since ${s_{n_k}}_k$ converges to $L$, we have that for all $epsiloninmathbb{R}^+$, there exists $Ninmathbb{N}^+$ and $ninmathbb{N}^+$ such that if $n_k geq k geq N$, then $|s_{n_k} – L| < frac{epsilon}{2}$.

Since ${s_{n_k}}_k$ is a subsequence of the monotone increasing sequence ${s_n}_n$, it follows that there exists $k’inmathbb{N}^+$ such that $n_{k’} geq k’ geq n$, which implies $s_n leq s_{n_{k’}}$ – because ${s_{n_k}}_k$ is increasing.

Subtracting by $s_{n_k}$ yields the following inequality:
begin{equation*}
s_n – s_{n_k} leq s_{n_{k’}} – s_{n_k}
end{equation*}
and finally:
begin{equation*}
|s_n – s_{n_k}| leq |s_{n_{k’}} – s_{n_k}|.
end{equation*}
But we know that ${s_{n_k}}_k$ is convergent, and thus Cauchy. So we can conclude that for all $epsiloninmathbb{R}^+$, there exists an natural number $N’$, for all natural numbers $n_{k’},n_k$ such that if $n_{k’},n_kgeq k’,kgeq N’$, then $|s_{n_{k’}} – s_{n_k}| < frac{epsilon}{2}$. Coupled with (i) we get that $|s_n – s_{n_k}| < frac{epsilon}{2}$.
begin{equation*}
|s_n – L| leq |s_n – s_{n_k}| + |s_{n_k} – L| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.
end{equation*}
${s_n}_n$ converges to $L$.