Does the series converge or diverge?

Question

Use OCT and $ln(n) < n^k$ to see if it converges or diverges.

$sum_(n=2)^oo ln(n)/n^(3/2)$

Instead of using the p-series theorem to solve this, I need to use the natural log theorem.

Theorem: The natural log function ln(x) is eventually below any positive power function $x^k$ . This means that ln(n) < n^k for any positive number k, if n is big enough.

2992 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-27T08:31:45

I think you actually need both:

$(1)$ we know that $lim_(x->oo) lnx/x^k = 0$ for any $k > 0$ , then also:

$lim_(n->oo) lnn/n^k = 0$

Choose now $k=1/4$ and $epsilon = 1$ . By the definition of the limit, we can find $N$ such that:

$n>N => lnn/n^(1/4) < 1$

$(2)$ For $n > N$ we have:

$lnn/n^(3/2) = lnn/n^(1/4)1/n^(5/4) < 1/n^(5/4)$

As $sum 1/n^(5/4)$ is a convergent series based on the $p$ -series test, by direct comparison also:

$sum_(n=2)^oo lnn/n^(3/2)$

is convergent.