Absolutely Convergent Series Converges

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!

• Yes! Your question is almost the definition of absolute convergence. – herb steinberg Apr 11 '19 at 3:04

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, $$\displaystyle\sum_{n=N+1}^\infty |a_n|<\varepsilon.$$

$$S_n =\sum_{k=1}^{n}|a_k|,$$ 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?