Question

Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Answer 1

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$. This means for any $\varepsilon>0$, there is some $N$ such that $|s_n-L|$ for all $n>N$. From this inequality we see that $|(s_n-L)-0|$, so it follows that $s_n-L\to0$.

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$, and we get the simpler inequality for free, $|s_n-L|$, so it follows that $s_n\to L$.