Shouldn't the harmonic series converge?

If a sequence converges in a metric space, it is Cauchy, and in $\mathbb{R}^k$ every Cauchy sequence converges.
Therefore, in $\mathbb{R}^k$ a sequence converges iff it is Cauchy.
Let $\{s_n\}$ be a sequence in $\mathbb{R}$ where each $s_n=\sum_{k=1}^na_k$. Therefore, by the above, every series converges iff $$\left | \sum_{k=m}^n a_k\right| <\epsilon$$ For a given $\epsilon >0$ and an integer $N$ such that $N\le m\le n$. If $n=m$ then the statement reduces to:
A series converges if and only if $$|a_n| < \epsilon$$ For a given $\epsilon >0$ and an integer $N$ such that $N\le n$.
This clearly cannot be (e.g Harmonic series). When does the equivalence become an implication.

• No it shall not. – Turing Jan 11 '17 at 16:14
• Maybe it should, but it doesn't want to... :-) – zipirovich Jan 11 '17 at 16:16
• Why are you capitalizing the word "For"? Those sentences need to be checked carefully. – littleO Jan 11 '17 at 16:23
• @AlanTuring I +1'd before I noticed your name. Now I want to +1 again. – Devsman Jan 11 '17 at 21:40
• @zipirovich lol! – Turing Jan 11 '17 at 22:53

In the definition of a Cauchy sequence, you have not gotten the order of the quantifiers correct. The partial sums of a series are Cauchy iff for all $\epsilon > 0$ there exists an $N$ such that for all $m,n \ge N$, $|\sum_{k=m}^{n} a_k| < \epsilon$. Notice the "for all $m,n$." You cannot just let $m=n$ and verify the condition for that case.

• But baby Rudin does, and that is where he gets one side of the implication: if the series converges then the terms go to zero. – Guacho Perez Jan 11 '17 at 16:28
• @GuachoPerez So Rudin says: :"If the series converges, the terms go to zero". However, it not the case that "If the terms go to zero, the series converges.". Of course, if you have shown that the series converges, you have for all $\epsilon$ and $N$ such that for all $n,m \ge N$ the partial sums are less than $\epsilon$. In that case, you can choose $n=m$ to see that $|a_n| < \epsilon$. – Hetebrij Jan 11 '17 at 16:46
• @Hetebrij is that precisely because of the fact that convergence implies Cauchy? – rb612 Mar 16 '18 at 8:13

It is true that a sequence in $\mathbb{R}$ converges if and only if it is Cauchy, and as such a series in $\mathbb{R}$ converges if and only if the sequence of partials sums $\{S_k\}_{k=0}^\infty$ is Cauchy. The problem is that you are not using the definition of Cauchy correctly. For a sequence $\{x_k\}_{k=0}^\infty$ it reads that for every $\epsilon >0$ there exists $K \ge 0$ such that if $m,n \ge K$, then $|x_n - x_m | < \epsilon$. Here one does not get to restrict to $n = m+1$, one must check all possible values of $m,n \ge K$. Note, though, that it is possible to restrict to $m \ge n \ge K$ in the definition.

Let's look at this for the harmonic series. In this case $S_k = \sum_{m=1}^k 1/m$. You are looking at $S_{k+1} - S_k = 1/(m+1)$, which indeed can be made arbitrarily small. In fact, to use the Cauchy definition we must consider arbitrary $k \ge \ell$ to get $$S_k - S_\ell = \sum_{m=\ell+1}^k \frac{1}{m},$$
and it is these partial sums of arbitrary length that cannot be made small.

If $\epsilon>0$ and $m\in\mathbb{N}$ then there is an $n\in\mathbb{N}$, $n>m$ such that

$$\sum_{k=m}^n\frac{1}{k}>\epsilon$$

Therefore, by the above, every series converges iff $$\left | \sum_{k=m}^n a_k\right| <\epsilon$$ for a given $\epsilon >0$ and an integer $N$ such that $N\le m\le n$.

What are $m$ and $n$ in this statement? They have not been introduced. This statement doesn't make sense.

Here's what you should have said.

The following two statements are equivalent:

1. The series $\sum_{k=1}^\infty a_k$ converges.
2. If $\epsilon > 0$, then there exists a positive integer $N$ such that if $m$ and $n$ are positive integers and $N < m \leq n$, then $\left | \sum_{k=m}^n a_k \right| < \epsilon$.

I think you're confusion is with the quantifiers. Your initial statements are correct, but to put everything in more specific terms the correct statement, with correct quantifiers is:

Let $\{s_n\}$ be a sequence in $\mathbb{R}$ where $s_n=\sum_{k=1}^na_k$. Then the series converges iff $\textit{for all } \epsilon > 0$ $\textit{there exists } N\in\mathbb{N}$ such that $\textit{for all } n\geq m > N$ we have that $$\left|\sum_m^na_k\right| < \epsilon$$

This does not hold for the harmonic series since $\sum_{k=N}^\infty\frac{1}{k}=\infty for all N$

It is true that a series converges if and only if the following is true:

For any $\epsilon > 0$, there is some $N$ such that $$\left\lvert \sum_{k=m}^n a_k\right\rvert < \epsilon$$ whenever $N\leq m \leq n$.

The statement in the box above implies this one:

For any $\epsilon > 0$, there is some $N$ such that $$\lvert a_n \rvert < \epsilon$$ whenever $N\leq n$.

The reason is that the statement in the first box says what happens for all $m,n$ such that $N\leq m \leq n$, including the pairs $m,n$ in which $m=n$.

On the other hand, the statement in the second box does not tell you what happens for all $m,n$ such that $N\leq m \leq n$, only what happens in a small subset of those cases. Therefore you cannot use the statement in the second box to conclude the statement in the first box.

A correct statement using the second box is,

A series $\{a_n\}$ converges only if for any $\epsilon > 0$, there is some $N$ such that $$\lvert a_n \rvert < \epsilon$$ whenever $N\leq n$.

What you did was to write "if and only if" when you should have written "only if".