At a glance, we can tell that the series converges if we use the handy fact that sum1/n^a converges if and only if a>1. That the given series converges then is just a matter if noticing that 1/(nsqrt(n)) = 1/n^(3/2).
However, let's see how we could prove this using a test for convergence or divergence.
Using the integral test:
This test states that if f(n) = a_n for all n in ZZ^+, then sum_(n=1)^oo converges if and only if int_1^oof(x)dx converges. To apply this, we let f(x) = 1/(xsqrt(x)) = x^(-3/2). Then
int_1^oof(x)dx = int_1^oox^(-3/2)dx
=[x^(-1/2)/(-1/2)]_1^oo
=-2[1/sqrt(x)]_1^oo
=-2(1/oo-1/1)
=2
As int_1^oof(x)dx converges and f(n)=1/(nsqrt(n)) at each positive integer n, sum_(n=1)^oo1/(nsqrt(n)) converges by the integral test (note that it does not necessarily converge to the same value as the integral).
As a side note, we could substitute in 1/x^a for 1/x^(3/2), perform the same test, and find that the integral converges if and only if a>1, verifying the shortcut mentioned in the start.
