Answer
The sequence $b_n=a_{n+1}$ also converges.
Work Step by Step
When $\lim\limits_{x \to \infty} f(x)$ exists, then the sequence $a_n=f(n)$ converges to the same limit, so we have:
$ \lim\limits_{n \to \infty} a_n=L$
We have:
$\lim\limits_{n \to \infty} (a_n -L)=0$
or, $\lim\limits_{n \to \infty} (a_n+a_{n+1} -a_{n+1} -L)=0$
or, $\lim\limits_{n \to \infty} (a_{n+1} - L) + \lim\limits_{n \to \infty} (a_n-a_{n+1})=0$
or, $\lim\limits_{n \to \infty} (a_{n+1} - L) =0$
We are given that: $a_{n+1}=b_n$
So, $\lim\limits_{n \to \infty} (b_n - L) =0 \implies \lim\limits_{n \to \infty} b_n = L$
We see that the sequence $b_n=a_{n+1}$ also converges.
