Given a sequence $a_n$ which converges to the limit $l$, how do I show the convergence of the sequence $b_n=a_{n+1}$ using $\epsilon$-definition of convergence of a sequence? I know that $b_n$ will also converge to the same limit $l$ and this is how I tried to show it. Given $\epsilon > 0$, since $a_n$ is convergent there exists some natural number $N$ such that- $$ |a_n-l|<\epsilon $$ $\forall n>N.$ Now replacing $n$ by $n+1$, we get- $$|a_{n+1}-l|<\epsilon$$ $\forall n>N-1$. So we can say that the sequence $b_n$ converges to $l$. Is this reasoning correct?
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Well it's automated won't it, $a_n$ converge to a limit $l$ Then suppose not and achive a contradiction!!! It works because $a_n$ is cauchy!!! – Oct 21 '22 at 19:16