If a series $\sum\limits_{k=1}^{\infty}a_{k}$ converges, then $(a_{k})\to 0$.

I want to go with the way Abbott was illustrating in his "Understanding Analysis." He uses the Cauchy Criterion for Series (Theorem 2.7.2) to prove this Theorem. Here is my attempt at what he was going for:

$\sum\limits_{k=1}^{\infty}a_{k}$ converges

$\implies[\forall\epsilon>0][\exists N\in\mathbb{N}][n>m\geq N\implies|a_{m+1}+\dots+a_{n}|<\epsilon]$.

Consider when $n=m+1$, then $|a_{m+1}-0|<\epsilon$.

Let $k=m+1$.

So,

$[\forall\epsilon>0][\exists N\in\mathbb{N}][k\geq N\implies |a_{k}-0|<\epsilon]$

$\implies (a_{k})\to 0$.

Solution 1:

More easily : if $S_n = \sum_{k=1}^n a_k$ then $a_n = S_n - S_{n-1} \to S - S = 0$ where $S = \sum_{k=1}^{\infty} a_k$

Solution 2:

Theorem. Let $\{a_n\}_{n\geq n_0}$ a real sequence. Necessary condition so that $\sum_{n=n_0}^{+\infty}a_n$ converges is that $a_n\to 0$ when $n\to +\infty$

Proof. Let $s_m=\sum_{n=n_0}^{m}a_n$ the sequence of the partial sums. We have that: $$s_{m+1}=\sum_{n=n_0}^{m+1}a_n$$ Now, we can calculate the limit: $$\lim_{m\to +\infty}s_{m+1}-s_m$$ Lemma. Let $\{c_n\}_{n\geq n_0}$. If $\lim_{n\to +\infty}c_n$ exist, then $\lim_{n\to +\infty}c_{n+1}$ and $\lim_{n\to +\infty}c_n$ are equal.

Thus: $$\lim_{m\to +\infty}s_{m+1}-s_m=0\implies \lim_{m\to +\infty}a_{m+1}=0$$

So, $a_n\to 0$ when $n\to+\infty$.

Note that is a necessary condition because $\sum_{n=1}^{+\infty}\frac{1}{n}$ diverges, but $\frac{1}{n}\to 0$. Instead, $\frac{1}{n^2}\to 0$ and $\sum_{n=1}{+\infty}\frac{1}{n^2}$ converges.