Let #f(x)=1/sqrt{2x-5}=(2x-5)^{-1/2}#. The improper integral #\int_{3}^{infty}f(x)\ dx# can be shown to diverge by direct calculation. First, do the indefinite integral #int (2x-5)^{-1/2}\ dx# by using the substitution: #u=2x-5#, #du=2\ dx# to write it as #1/2 int u^{-1/2}\ du=u^{1/2}+C=sqrt{2x-5}+C#.
Now write the improper integral as a limit:
#int_{3}^{infty}f(x)\ dx=lim_{b->infty}int_{3}^{b}(2x-5)^{-1/2}\ dx=lim_{b->infty} sqrt{2x-5}|_{x=3}^{x=b}#
#=lim_{b->infty}(sqrt{2b-5}-1)#
This last limit doesn't exist (some would say it "diverges to #infty#").
Since #f(x)>0# for all #x\geq 3# and since #f# is decreasing on the interval #x\geq 3#, it follows from the integral test that the series #sum_{n=3}^{infty}1/sqrt{2n-5}# diverges.
Since the series in question #sum_{n=1}^{infty}1/sqrt{2n-5}# just contains two more terms at the beginning, it follows that #sum_{n=1}^{infty}1/sqrt{2n-5}# diverges as well. And actually, the first two terms of this series are complex numbers since #2n-5<0# when #n=1# and #n=2#. So maybe you meant something different than your original question anyway.
