real analysis - Suppose that the series $sum a_n$ converges. Prove that $limlimits_{ntoinfty} a_n = 0$.

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.

Answer

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 $n > N_2$:

$$\begin{array}{rcll} |A_{n+1} - L| &<& \dfrac12 \varepsilon_2 & \text{assumption} \\ |A_n - L| &<& \dfrac12 \varepsilon_2 & \text{assumption} \\ |A_{n+1} - A_n| &<& |A_{n+1} - L| + |A_n - L| & \text{triangle inequality} \\ &<& \dfrac12 \varepsilon_2 + \dfrac12 \varepsilon_2 \\ &=& \varepsilon_2 \\ |a_n| &<& \varepsilon_2 & \text{conclusion} \end{array}$$