Using the integral test, how do you show whether #(1/sqrt (n+1))# diverges or converges?

1 Answer
Jun 11, 2015

I would assume that this goes from #n = 1# to #oo#? The domain must be #[k,oo]# in order for this to work. If so, I've already answered this here:
http://socratic.org/questions/using-the-integral-test-how-do-you-show-whether-sum-1-sqrt-n-1-diverges-or-conve#151880

Overall, it diverges, because the integral evaluates as #oo#. The gist of it is, integrate it replacing #n# with #x#, and show that it will result in #oo# when evaluated from #1# to #oo#.

You should also know that this is a "p-series":

#sum_(n = 1)^(oo) 1/((npmk)^p)#
where #k in RR# (#k# is in the set of real numbers)

If #p > 1#, the sum will converge, whereas if #p <=1#, the sum will diverge. When #p = 1#, it's called the harmonic series