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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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2 Answers 2

up vote 7 down vote accepted

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

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3  
That's fine. The standard argument goes $a_n = \sum_{i}^n a_i-\sum_{i}^{n-1}a_i\to L-L=0$ as $n \to \infty$. –  Peteris Feb 10 '12 at 23:36
    
Thanks for the feedback –  ItsNotObvious Feb 11 '12 at 0:52

That the series converges means that the sequence of partial sums

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

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

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