How to do comparison test for #sum 1 / (sqrt(n))# n=1 to #n=oo#?

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Answer 1 · 2015-05-15T22:29:50

The series diverges.

Let's begin with an inequality :

#sqrtn <= n, #, with #n >=0#

#1/sqrtn>= 1/n#, with #n>=1#

#sum_(n=1)^oo1/sqrtn>=sum_(n=1)^oo1/n#

#sum_(n=1)^oo1/n# is the harmonic series and it diverges.

Therefore, by comparison test, #sum_(n=1)^oo1/sqrtn# diverges.