If we can derive that $\{x_{f(n)+1}\}$ converges to $a$ from that $\{x_{f(n)+1}\}$converges to $a$

Question

Just like in the title: if a subsequence $\{x_{f(n)}\}$ of $\{x_n\}$ converges to $a$, and based on the fact that $\{x_{f(n)}\}$converges to $a$, we can derive that $\{x_{f(n)+1}\}$, $\{x_{f(n)+2}\}$,$\{x_{f(n)+3}\}...$ also converge to $a$, is it true that $\{x_n\}$ converges to a? (It seems very true because it seems we can cover the whole sequence using (maybe infinite) converged subsequences)

Answer 1

This is wrong. A Counter example: $x_{n}=0$ for $n=t^2$, $x_n=1$ otherwise. And let $f(s)=s^2+1$.

Answer 2

$x_{3n}=0, x_{3n+1}=0, x_{3n+2}=1$