Does the series #sum_{n=1} ^oo (2)/(n^2sqrtn)# converge or diverge?

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Answer 1 · 2018-01-15T19:39:22

Converge

Use the comparison test.

We know that (from the Riemann-zeta function):

#sum_(n=1)^oo2/n^2=2sum_(n=1)^oo1/n^2=2zeta(2)=2*pi^2/6=pi^2/3#

So

#sum_(n=1)^oo2/n^2# converges.

We also have that for all values of #n#:

#2/(n^2sqrt(n))lt2/n^2# therefore:

#sum_(n=1)^oo2/(n^2sqrt(n)# must also converge by the comparison test.