My question is as follows

> 
> Let $(b\_n)$ be the sequence obtained from $(a\_n)$ by deleting terms so that infinitely many remain. Show that if $(a\_n) \to \alpha$ then $(b\_n) \to \alpha$.
> 
> 
> 

A previous question asks the same but defines $b\_n = a\_{n+1}$ and so is a "base case" of the above question. In that question, I simply stated that since $(a\_n) \to \alpha$, we have that, given an $\varepsilon > 0$, there exists an $N \in \Bbb N$ such that

$$\lvert a\_n - \alpha\rvert < \varepsilon,\ \forall n > N$$

and that if $b\_n = a\_{n+1}$ then we simply define $N^\prime = N - 1$ and have that, given $\varepsilon > 0$, there exists an $N'$, namely $N' = N-1$ such that

$$\lvert b\_n - \alpha\rvert < \varepsilon,\ \forall n > N'$$

and hence have the result that $\lim b\_n = \alpha$.

Given that the above is somewhat correct (hopefully), is it wise then for this question to define $b\_n = a\_{n+k}$ and simply set $M = N - k$ for a proof, or does this use an unnecessary assumption? (That is, that $b\_n$ is a copy of $a\_n$ shifted along $k$ spaces, rather than that we can simply remove some terms of $a\_n$).