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

Question

Emma Aldridge · Accepted Answer

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.

Charles Arocha · Answer

undefined To apply the comparison test for the series $ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} $, we need to find a series whose terms are always less than or equal to the terms of the given series, and whose convergence or divergence we know.

Given that $ \frac{1}{\sqrt{n}} $ is positive for all $ n $, we can compare it to the series $ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where $ p > 0 $. Specifically, we can choose $ p = \frac{1}{2} $ since $ \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} $.

We know that the series $ \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} $ is a p-series with $ p = \frac{1}{2} $, and it converges because $ p > 0 $.

Therefore, by the comparison test, since $ \frac{1}{\sqrt{n}} $ is always less than or equal to $ \frac{1}{n^{1/2}} $ for all $ n $, and $ \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} $ converges, then $ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} $ also converges.