Quote by morphism
Just because (a_n  a_{n+1}) > 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n)  log(n+1) = log(n/(n+1)) > log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't.
It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge  it goes around the unit circle in the complex plane. On the other hand,
[tex]a_n  a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1  \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex]

Okay, how about [tex]a_na_{n1} \le k \left( \frac{1}{n}\frac{1}{n1} \right) [/tex] for all n greater then M