$ \sum_{n=1}^{\infty} a_n \ $ converges then show that $ \ |a_n|<1 \ $
Answer:
If not let $ \ |a_n|>1 \ $
Then the partial sum is $ \ s_n=|a_1|+|a_2|+....+|a_n| \ $
Let $ \ n>m \ $, then
$ |s_n-s_m|=|a_m|+|a_{m+1}|+........+|a_n|> (n-m) \ $
since $ \ n>m $ , we have $ \ n-m \geq 1 \ $
This shows that
$ |s_n-s_m| \nrightarrow \ 0 $
This implies $ s_n \ $ is not Cauchy .
Therefore $ \ \sum a_n \ $ diverges.
Which is a contradiction.
Thus we must have $ \ |a_n|<1 \ $
Am I right so far?
If you want to show that $|a_n|<1$ after a certain stage then NO your answer does not hold because the logical negation of: $$ \exists N\in \mathbb{N}, \forall n\geq N, |a_n|<1 $$ is not: $$ \exists N\in \mathbb{N}, \forall n\geq N, |a_n|>1 $$ but rather: $$ \forall N\in \mathbb{N}, \exists n\geq N, |a_n|>1 $$ In other words, if you are going reductio ad absurdum (which is perfectly legitimate), you may use the fact that if there were infinitely many $\Phi(n)$ so that $|a_{\Phi(n)}|>1$ then since $\sum_n a_n$ converges, $\lim_{n\to \infty} a_n = \lim_{n\to \infty} a_{\Phi(n)}=0$. The contradiction follows.