# Theorem 3.22 from baby Rudin

$\sum a_n$ converges if and only if for every $\varepsilon >0$ there is an integer $N$ such that $$\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$$ if $m\geqslant n\geqslant N$.

In particular, by taking $m=n$ above inequality becomes $$|a_n|\leqslant \varepsilon \quad(n\geqslant N).$$

In other words: If $\sum a_n$ converges, then $\lim_{n\to \infty}a_n=0$

The condition $a_n\to 0$ is not, however, sufficient to ensure convergence of $\sum a_n$. For instance. the series $\sum \frac{1}{n}$ diverges.

Reading all this I have one question.

If for every $\varepsilon >0$ there is an integer $N$ such that $\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$ if $m\geqslant n\geqslant N$ then $\sum a_n$ converges.

If we put here $m=n$ why can not conclude that $\sum a_n$ converges? Where is the mistake?

• It has to be true for EVERY $m\geq n\geq N$. Taking $m=n$ is only a special case, so it doesn't give enough restriction on the sequence $(a_n)_{n\in \mathbb{N}}$. – Arnaud D. Aug 16 '15 at 16:26
• Can you give some simple example? It would be great! – ZFR Aug 16 '15 at 16:41
• Well the best example is already in your question : the harmonic series $\sum\frac{1}{n}$. To expand a little bit : take $\epsilon =\frac{1}{2}$ and let $N\in \mathbb{N}$. Then choose $n=2^p\geq N$ and $m=2^{p+1}$. For all $k$ with $m\geq k\geq n$ you have $\frac{1}{k}\geq \frac{1}{m}=2^{-(p+1)}$, so $\sum_{k=n}^m\frac{1}{k}\geq (2^{p}+1)2^{-(p+1)}> \frac{1}{2}=\epsilon$. So even though the condition holds when you choose $m=n$, it doesn't necessary hold in other cases, and thus the series doesn't need to converge. – Arnaud D. Aug 16 '15 at 16:55

## 2 Answers

Because statement is

"$\sum a_n$ converges if and only if for every $\varepsilon >0$ there is an integer $N$ such that $$\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$$ if $m\geqslant n\geqslant N$."

Notice that for $\sum a_n$ to converge, $$\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$$ should happen for all $m\ge n\ge N$.

Choosing $m=n$ is just one case.

In your example, $\sum\frac1n$ does not converge as for every given $\varepsilon$, you cannot find an $N>0$ such that $$\left| \frac1n+\frac1{n+1}+\dots \frac1m \right| < \varepsilon$$ for all $m\ge n\ge N$

Although you can find an $N$ for the case $m=n$, i.e. for every given $\varepsilon$, we can find an $N$ such that $$\left| \frac1n \right|< \varepsilon$$ for all $n \ge N$, by archimedean property.

The mistake is that $$\sum_{k=m}^m a_k = a_m$$ and you are saying nothing about the series $\sum_k a_k$. You are only looking at the single term $a_m$.

• Why? We have that $|a_m|<\varepsilon$ for $m\geqslant N$. We have that: If some statement $P$ is true for EVERY $m\geqslant n\geqslant N$ then statement $N$ is true. Why we can conclude from above that if statement $P$ is true for every $m=n\geqslant N$ then $N$ is true? – ZFR Aug 16 '15 at 16:34
• In you phrasing, $Nˆ$ is both a number and statement. You have to choose! – Bernard Aug 16 '15 at 16:36
• Choose $a_k=1/k$ for every $k$. given $\epsilon>0$, it is false that there exists $N>0$ such that $\sum_{k=m}^n a_k>\epsilon$ for all $n$ and $m$ larger than $N$. However, it is true that $\lim_{k \to \infty} a_k=0$. – Siminore Aug 16 '15 at 16:50
• Dear Bernard! Sorry! Let $N$ be number but the second statement be $T$. Example with harmonic series is harder to me maybe there is some more simple example? – ZFR Aug 16 '15 at 17:33