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|<\varepsilon` for all
`n>N`. From this inequality we see that
`|(s_n-L)-0|<\varepsilon`, 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|<\varepsilon` for all large enough
`n`, and we get the simpler inequality for free,
`|s_n-L|<\varepsilon`, so it follows that
`s_n\to L`.