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.
