Suppose that the series $ \sum a_n$ converges. Prove that $\lim\limits_{n\to\infty} a_n = 0$. Suppose that the series $\sum a_n$ converges. Prove that $\lim\limits_{n\to\infty}
a_n = 0$.
Not sure how to do this my attempt:
WTS:$\forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - 0| < \epsilon$
Let $\epsilon > 0$ be arbitrary
Choose N such that $|a_n - 0| < \epsilon$
Suppose $n > N$, then $|a_n - 0| < \epsilon$ 
I am not sure. 
 A: Suppose that $\sum\limits_{n=1}^{\infty}{a_n}=M$ then we have that $M=\lim_{n \to \infty}({a_n} + \sum\limits_{k=1}^{n-1}{a_k})=\lim_{n\to\infty}{a_n}+\lim_{n\to\infty}{\sum\limits_{k=1}^{n-1}{a_n}}=\lim_{n\to\infty}a_n+M$. Hence, we have that $a_n \to 0$. 
A: Let $A_n := \displaystyle \sum_{k=1}^n a_k$. Note that $a_n = A_n - A_{n-1}$.
Then, $\displaystyle \sum_{k=1}^\infty a_k := \lim_{n\to\infty} A_n = L$.
Now, since the sum converges:
$$\forall \varepsilon_1 > 0: \exists N_1 \in \Bbb N: \forall n > N_1:|A_n-L| < \varepsilon_1$$
Now, we need to prove that $a_n$ goes to zero:
$$\forall \varepsilon_2 > 0: \exists N_2 \in \Bbb N: \forall n > N_2:|a_n| < \varepsilon_2$$
To prove that, let $\varepsilon_1 = \dfrac12\varepsilon_2$. Now, we have $N_1$ from the assumption. Let $N_2 = N_1+1$. For any $n > N_2$:
$$\begin{array}{rcll}
|A_{n+1} - L|  &<& \dfrac12 \varepsilon_2 & \text{assumption} \\
|A_n - L|  &<& \dfrac12 \varepsilon_2 & \text{assumption} \\
|A_{n+1} - A_n|  &<& |A_{n+1} - L| + |A_n - L| & \text{triangle inequality} \\
&<& \dfrac12 \varepsilon_2 + \dfrac12 \varepsilon_2 \\
&=& \varepsilon_2 \\
|a_n| &<& \varepsilon_2 & \text{conclusion}
\end{array}$$
A: Let $$L=\sum^{\infty}_{n=1}a_{n}=\lim _{n\to \infty}\sum^{n}_{k=1}a_{k}=L$$
Then, $A_{n}$ is from Cauchy, that is:
$$\forall \epsilon >0, \exists n_{0}\in \mathbb{N}/ n,p\geq n_{0}\rightarrow |A_{p}-A_{n}|<\epsilon$$
In particular, for $$p=n+1\rightarrow |A_{n+1}-A_{n}|<\epsilon \rightarrow \left | \sum^{n+1}_{k=1}a_{k}-\sum^{n}_{k=1}a_{k}\right|<\epsilon \rightarrow |a_{n+1}|<\epsilon\rightarrow$$
Taking $$n_{1}=n_{0}+1\rightarrow \forall \epsilon >0, \exists n_{0}\in \mathbb{N}/ n\geq n_{1}\rightarrow |a_{n}-0|<\epsilon \rightarrow \lim _{n\to \infty}a_{n}=0$$
