Let $\sum_{n=1}^\infty a_n$ an converges absolutely, $(a_n)$ being a series. Prove for every $\epsilon >0, \exists N \in \mathbb{N}$ s.t. $\sum_{k=N+1}^\infty |a_n| < \epsilon$.
So if it converges absolutely, then $\sum_{n=1}^\infty |a_n|=L$ and $\sum_{k=N+1}^\infty |a_n| = L- \sum_{n=1}^N |a_n| < \epsilon$.
Find $n$ such that $L- \sum_{n=1}^N |a_n|<\epsilon'$ since this is a Cauchy sequence if the series of partial sums converges. AM i on the right track thank!
On
$S_n =\sum_{k=1}^{n}|a_k|,$ i.e. is Cauchy.
For $\epsilon >0$ there exists a $n_0$ s t. for $m \ge n \ge n_0$.
$|S_m-S_n| \lt \epsilon$,
$\sum_{k=n+1}^{m}|a_k| <\epsilon.$
Set $n=n_0$, and take the limit $m \rightarrow \infty$:
$\sum_{n_0+1}^{\infty}|a_k| \le \epsilon$.
Note the $\le$ sign in above expression . If you want a < sign, what do you change?
Since the series converges absolutely, then the sequence of partial sums $$S_k:=\displaystyle\sum_{n=1}^k |a_n|$$ is a convergent sequence converging to $S:=\displaystyle\sum_{n=1}^\infty |a_n|$, that is, for every $\varepsilon>0$, there exists an $N\in\mathbb{N}$ such that for every $k\geq N$, $|S_k-S|<\varepsilon.$ But $|S_k-S|=\displaystyle\sum_{n=k+1}^\infty|a_n|$. So in particular, $\displaystyle\sum_{n=N+1}^\infty |a_n|<\varepsilon.$