<s>[INST] Prove that $a_n$ is a Cauchy sequence, if $|a_{n+1} - a_n| < \frac{1}{n(n+1)}$ Obviously, $\frac{1}{n(n+1)}$ converges to $0$, as $n \to \infty$
But it doesn't say anything to me, since, for example, $a_n = \sqrt{n}$ diverges, but $|a_{n+1} - a_n| \to 0$
So I need to look at this problem from another point of view.
Sequence is Cauchy, if $$(\forall \varepsilon >0)(\exists N \in \mathbb{N})(\forall n \geq N)(\forall m \geq N): |a_n - a_m| < \varepsilon$$
Now considering some inequalities:

*

*$|a_{n}   - a_{n+1}| < \frac{1}{(n)  (n+1)} = \frac{1}{n}   - \frac{1}{n+1}$

*$|a_{n+1} - a_{n+2}| < \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}$

*$|a_{n+2} - a_{n+3}| < \frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$ etc.

Adding them:
$|a_n - a_{n+3}| \leq |a_{n}   - a_{n+1}| + |a_{n+1} - a_{n+2}| + |a_{n+2} - a_{n+3}| < \frac{1}{n}   - \frac{1}{n+1} + \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+2} - \frac{1}{n+3} = \frac{1}{n} - \frac{1}{n+3}$
And that makes me assume, that $|a_n - a_m| < \frac{1}{n \times m}$ (if it is a correct assumption, induction is needed to prove it, I guess)
Hence, in order to make $|a_n - a_m| < \varepsilon$, we need to ensure that $n \times m > \frac{1}{\varepsilon}$
But I would like to have only $n$ on the left side.
So, besides the need to prove that $a_n$ is a Cauchy sequence, I have four additional questions:

*

*Is my reasoning correct?

*If it is, then do we need to get rid of $m$ (since $m$ is not a fixed number).

*If we do, how to do that?

*If we are talking about basic Cauchy sequences, is tag "sequences-and-series" appropriate?

Thank you
Edit:
instead of $|a_n - a_m| < \frac{1}{n \times m}$ there should be $|a_n - a_m| < \frac{1}{n} - \frac{1}{m}$
 [/INST] For $n<m$, write
$$
a_n-a_m = \sum_{k=n}^{m-1}a_k-a_{k+1},
$$
so that the triangle inequality yields
$$
|a_n-a_m| \leqslant \sum_{k=n}^{m-1}|a_k-a_{k+1}|.
$$
From what you have done, you should find
$$
|a_n-a_m| \leqslant \frac{1}{n}-\frac{1}{m} \leqslant \frac{1}{n},
$$
and the result should come up easily now.
This is the so called trick of telescoping series, and getting rid of $m$ is easier than you thought here.
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