I apologize, I'm so new at proofs. This should be very easy, but I'm not sure how to say it. This is what I've come up with. Is it sufficient? Is there another way I should be looking at this?
Thanks in advance!
I apologize, I'm so new at proofs. This should be very easy, but I'm not sure how to say it. This is what I've come up with. Is it sufficient? Is there another way I should be looking at this?
Thanks in advance!
I don't usually prove statements by using contradictions, but here's how I'd solve this:
Given $\displaystyle \sum_{n=1}^\infty a_n<\infty,$ then $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n a_k$ exists, and so does $\displaystyle \lim_{n\to\infty}\sum_{k=1}^{n-1} a_k,$ hence $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\left( \sum\limits_{k=1}^{n}{a_{k}}-\sum\limits_{k=1}^{n-1}{a_{k}} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( \sum\limits_{k=1}^{n-1}{a_{k}}+a_{n}-\sum\limits_{k=1}^{n-1}{a_{k}} \right)=0,$ then $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,a_{n}=0,$ as claimed.
Let $\displaystyle \sigma_m=\sum_{n=1}^{m}a_n$. Since, $\displaystyle \lim\text{ }\sigma_m$ exists (i.e. it converges) we have that it is Cauchy. Thus, for every $\displaystyle \varepsilon>0$ there exists an $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N<\ell\leqslant k\implies\left|\sigma_k-\sigma_\ell\right|=\left|\sum_{n=1}^{\ell}-\sum_{n=1}^{k}\right|=\left|\sum_{n=\ell}^{k}a_n\r ight|<\varepsilon$. Taking $\displaystyle \ell=k$ we see that for every $\displaystyle \varepsilon>0$ there exists an $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N<\ell\implies \left|\sum_{n=\ell}^{\ell}a_n\right|=\left|a_\ell\ right|<\varepsilon$. Therefore, $\displaystyle a_\ell\to0$.