If $a_n \ge 0$ and $a_n \to 0$, then $\sum a_n$ converges

Question

If $a_n \ge 0$ for all $n$ and $a_n \to 0$, then $\sum_{n=1}^\infty a_n$ converges

I recently answered a question different from this one, but found that my methodology could be used to prove the above result, which I know to be untrue (for example, take the harmonic sum). I'm just wondering where exactly my proof fails:

Checking the Cauchy Criterion for the sum:

$$|\sum_{n = 1}^k a_n - \sum_{n=1}^m a_n| = |\sum_{n=m}^k a_n| \le \sum_{n=m}^k |a_n|$$

utilizing the triangle inequality and arbitrarily letting $k > m$.

Since $a_n \to 0$, $\forall \epsilon > 0, \exists N$ such that $n > N \implies |a_n| < \frac{\epsilon}{k - m + 1}$ for any integers $k, m$ with $k > m$. Let $k, m > N$. Then,

$$\sum_{n=m}^k |a_n| < \epsilon$$

So that by the Cauchy criterion, the sum converges.

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The logic here seems pretty good to me, honestly. I'm assuming that it fails by defining $|a_n| < \frac{\epsilon}{k-m+1}$ because we assume $k, m$ to be constants in this definition, yet in the Cauchy criterion for the sum, they are variable.

hence, saying $|a_n| < \frac{\epsilon}{k-m+1}$ allows $\sum_{n=m}^r |a_n| < \epsilon$ only for $r \le k$, so that we cannot in general say it holds for all values $k, m > N$.

Is this the correct rationale? Could someone shed some (more intuitive) light on it if so, because the rationale I have provided is still a little bothersome to be, for it really is true that $|a_n| < \frac{\epsilon}{k-m+1}$ for any choice in $k, m > N$ with $k > m$, so it almost seems as though $k, m$ being variable would be okay.

Answer 1

The problem is in the quantifiers that are implicit in the statement you are making. What you have is that for all $\epsilon>0$ and all integers $k,m$ with $k>m>0$, there is an $N$ such that if $n>N$, then $|a_n|<\epsilon/(k-m+1)$. Of course, $N$ depends in general on $k,m$ and there is no way to choose it so that you can further require that $k,m$ are arbitrary numbers larger than $N$.

As a piece of advice, it is good to get used to listing all your quantifiers at the beginning of the statements, to avoid confusions such as the one in this case. This is tricky at first, since it is quite common in informal mathematical discourse to have the quantifiers at the end. From context, it is usually clear what we mean, but the practice leads to confusion (as you have experienced here).

Answer 2

The mistake is in the sentence "Since... " just before your second displayed equation. You probably understand that your $N$ depends on $\epsilon$, but it also depends on your $k$. If what you wrote were really true, you would have $|a_n|< \epsilon/(k-m+1)$ for all $k$ sufficiently large, which already implies $a_n= 0$, which is false even in some cases where $\sum a_n$ does converge.