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

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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 '12 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. $\endgroup$ – CameronJWhitehead Sep 13 '14 at 22:15
  • $\begingroup$ @CameronJWhitehead Good observation. $\endgroup$ – Erdogan CEVHER Aug 6 at 12:48
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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? $\endgroup$ – james black Apr 11 at 2:43
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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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