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.

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