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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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3 Answers 3

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

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    $\begingroup$ 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 $n \to \infty$. $\endgroup$
    – Peteris
    Feb 10, 2012 at 23:36
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    $\begingroup$ I know that this is hugely obvious but this is the beginning of analysis and its important to be annoying sometimes. In your second line you expand the limit into two limits, this is only true if the sequence $A_n$ converges. Which you have not proved, so I don't like this proof. I know that it is obvious, but without it this proof is incomplete. We know the partial sums converge, but we don't a priori know that the sequence of terms converges. If you prove first that it converges then I really like your proof. Sorry for such a huge comment. $\endgroup$ Sep 13, 2014 at 22:15
  • $\begingroup$ @CameronJWhitehead Good observation. $\endgroup$ Aug 6, 2019 at 12:48
  • $\begingroup$ @CameronJWhitehead If you formulate the proof like Peteris did, your problem disappears :) $\endgroup$ Oct 27, 2021 at 12:56
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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 $$\lvert a_n\rvert=\lvert s_n-s_{n-1}\rvert<\varepsilon\qquad\text{for all $n> N$.}$$

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  • $\begingroup$ third line, you mean (sn) instead of (sn)n? $\endgroup$ Apr 11, 2019 at 2:43
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    $\begingroup$ Sorry to make a comment on such an old thread. I understand the notion behind using the cauchy criterion condition, but I'm confused as to how this shows that the sequence converges to 0 $\endgroup$ Mar 4, 2020 at 21:03
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    $\begingroup$ @NeilsonsMilk, by definition, $(a_n)$ converges to zero iff for each $\varepsilon>0$ there is an $N\in\mathbb N$ such that $|a_n|<\varepsilon$ for all $n>N$. $\endgroup$
    – Carsten S
    Mar 5, 2020 at 9:56
  • $\begingroup$ @CarstenS oh right, because the condition is $|a_n-a|$ $<$ $\epsilon$ where $a = lim (a_n)$ $\endgroup$ Mar 7, 2020 at 17:10
  • $\begingroup$ @NeilsonsMilk, ah, it did not even occur to me that this involves a step. See, where I learned mathematics, it is not unusual to first define when a sequence converges to zero (and we have a word for those sequences, Nullfolge), and only then when a sequence converges to an arbitrary number, by considering the difference. $\endgroup$
    – Carsten S
    Mar 11, 2020 at 20:20
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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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