I have to prove the following:
Let $(a_n)$ be a sequence such that $(a_n)<1$. Prove or disprove:
If $(\frac{1} {1-a_n} )$ is bounded from above, then $\sup(a_n) \ne 1$.
I was thinking that I can prove the statement with the fact that if the sequence is bounded from above then $(\frac{1} {1-a_n} )<M$, by definition.
Now assume in contradiction that $\sup(a_n) =1$. Since $(a_n)<1$, $(\frac{1} {1-a_n} )>0$.
From above we come the a conclusion that $(a_n)<0$ which means the the supremum isn't 1.
Is my solution right? I know formality needs some work.
Hint: If $$ \frac{1}{1-a_n}<M, $$ then since $1-a_n>0$, we have $$ \frac{1}{M}<1-a_n, $$ which gives $$ a_n<1-\frac{1}{M} $$