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Following the guidelines suggested in this meta discussion, I am going to post a proposed proof as an answer to the theorem below. I believe the proof works, but would appreciate any needed corrections.

**Theorem** If a series $\sum_{n=1}^{\infty}a_n$ of real numbers converges then $\lim_{n \to \infty}a_n = 0$

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**Proof** If the series converges to the number $L$, this means that the sequence of partial sums converges to $L$, that is,

$$

\lim_{n \to \infty} \sum_{k=1}^n a_k = L.

$$

But,

$$

\lim_{n \to \infty} \sum_{k=1}^n a_k = \lim_{n \to \infty}\left( a_n + \sum_{k=1}^{n-1} a_k \right) = \lim_{n \to \infty}a_n + \lim_{n \to \infty} \sum_{k=1}^{n-1} a_k,

$$

however, as $n \rightarrow \infty$, the partial sum

$$

\sum_{k=1}^{n-1} a_k

$$

also converges to $L$. Therefore, the second equation can be rewritten as

$$

L = \lim_{n \to \infty} a_n + L \implies \lim_{n \to \infty}a_n = 0

$$

$\square$

That the series converges means that the sequence of partial sums

$$s_n=\sum_{k=1}^n a_k$$

converges. It follows that $(s_n)_n$ is a Cauchy sequence.

Now let $\varepsilon>0$. Since $(s_n)$ is a Cauchy sequence, there is an $N$ such that

$\lvert s_n-s_m\rvert<\varepsilon$ for all $m,n\ge N$. In particular

$$\lvert a_n\rvert=\lvert s_n-s_{n-1}\rvert<\varepsilon\qquad\text{for all $n> N$.}$$

Another view of this may be useful.

First, recall a basic fact that if $a_n$ is a convergent sequence of numbers, then the sequence $b_n = a_{n+1} – a_n$ converges to $0$. This is easy to prove and does not require the notion of a Cauchy sequence.

Therefore, if the partial sums $s_n$ are convergent, then $b_n = s_{n+1} – s_n$ converges to $0$. But the terms of this sequence are easily seen to be $b_n = a_{n+1}$. Hence $a_n \rightarrow 0$.

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