Quote Originally Posted by buckeye1973 View Post
Hmm.. How about this..

Take as a given that $\displaystyle \left\{1 - \frac{\log n}{n}\right\}$ is an increasing sequence with $\displaystyle \lim_{n \to \infty} \left(1 - \frac{\log n}{n}\right) = 1$. (It is true: see here.)

For $\displaystyle |r| < 1$ we know that

$\displaystyle \sum_{n=0}^\infty r^n = \frac{1}{1-r}$

but that same sum diverges for any other r.

Let $\displaystyle r \in (0,1)$. Then, by definition of limit, there exists $\displaystyle N \in \mathbb{N}$ such that for each $\displaystyle n \geq N$, $\displaystyle \left(1 - \frac{\log n}{n}\right) > r$. Thus for any $\displaystyle r \in (0,1)$, there are an infinite number of terms such that
$\displaystyle \left(1 - \frac{\log k}{k}\right)^k > r^k$. Hence, $\displaystyle \sum\left(1 - \frac{\log n}{n}\right)^n$ is greater than any (monic) convergent geometric series, and is therefore divergent.

Anyone? I'm afraid that this post has had so many replies that nobody wants to read it, but I haven't got a usable answer. Plato gave a good hint maybe, but I don't follow it..