# Show that a sequence $\{s_n\}$ converges to $L$ if and only if the sequence $\{-s_n\}$ converges to $-L$.

I understand that both parts of this biconditional must be proven. If I assume that a sequence $\{s_n\}$ converges to $L$,then for every for every $ϵ>0$, there is some integer N where $|s_n-L|<ϵ$ whenever n

• whenever $n < N$? – 3x89g2 Feb 13 '16 at 2:29
• I think it's pretty simple if you use $|-s_n-(-L)|=|s_n-L|$. – Alex Mathers Feb 13 '16 at 2:31

You can use the more general result (which is instructive to prove, and I urge you to do so) is that if $f$ is a continuous function, and if $s_n \to L$, then $f(s_n) \to f(L)$, as long as $L$ is in the domain of $f$. Since $f(x) = -x$ is continuous (proof?), we have that:
$$s_n \to L \implies -s_n \to -L.$$Conversely: $$-s_n\to -L \implies -(-s_n)\to -(-L),$$in other, words, $s_n \to L$, as wanted.