Question

How do i find the convergence or divergence of this series? `sum` from 1 to infinity of `1/n^lnn`

Answer 1

It converges

Explanation:

Consider the series `sum_(n=1)^oo1/n^p`, where `p>1`. By the p-test, this series converges.

Now,
`1/n^ln n<1/n^p` for all large enough `n` as long as `p` is a finite value.

Thus, by the direct comparison test, `sum_(n=1)^oo1/n^ln n` converges.

In fact, the value is approximately equal to `2.2381813`.