Convergance using Cauchy Criterion

Question

let $$\sum_{k=1}^n \frac{1}{n+k}$$ The sum is convergent since it satisfies the Cauchy criterion: $$|z_{n+p} - z_n| = \left|\sum_{k=n+1}^{n+p} \frac{1}{n+k}\right| \le \sum_{k=n+1}^{n+p}\frac{1}{n+n+1} = \frac{p}{2n+1} \to 0$$ for $n \to \infty$.

Can someone explain why and if this is true?

Answer 1

You can analyse this case by elementary means. You can simply note that, since $$ \frac{n}{n+n}\leq\underbrace{\sum_{k=1}^n \frac{1}{n+k}}_{a_n} \leq \frac{n}{n+1}, $$

the sequence $(a_n)$ in bounded ($\frac 12 \leq a_n \leq 1$). The sequence is also increasing and thus convergent.

Answer 2

The Cauchy criterion is for series, first of all, and this is not a series, because there are terms in your sum for $n$ that are not in your sum for $n+1.$

That said, even for series, just because, for fixed $p,$ $\sum_{k=n+1}^{n+p} |a_k|\to 0$ as $n\to\infty$ we can not conclude the series is convergent, because $a_k=\frac{1}{k}$ has similarly:

$$\sum_{k=n+1}^{n+p} \frac 1k\leq \frac p{n+1}\to 0.$$

Yet we know $\sum_{k=1}^\infty \frac1k$ diverges.

The error in this argument is really a complete misunderstanding of the Cauchy criterion.