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

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

For |r| < 1 we know that

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

but that same sum diverges for any other r.

Let r \in (0,1). Then, by definition of limit, there exists N \in \mathbb{N} such that for each n \geq N, \left(1 - \frac{\log n}{n}\right) > r. Thus for any r \in (0,1), there are an infinite number of terms such that
\left(1 - \frac{\log k}{k}\right)^k > r^k. Hence, \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..