How do you test the series #Sigma rootn(n)/n# from n is #[1,oo)# for convergence?

You can test the convergence of the series ( \sum \frac{\sqrt{n}}{n} ) from ( n = 1 ) to infinity using the limit comparison test.

First, observe that ( \frac{\sqrt{n}}{n} ) can be simplified to ( \frac{1}{\sqrt{n}} ).

Then, consider the series ( \sum \frac{1}{\sqrt{n}} ). This series is a p-series with ( p = \frac{1}{2} ), and it is known that p-series with ( p < 1 ) converge.

Now, take the limit of the ratio of the given series and the series ( \sum \frac{1}{\sqrt{n}} ) as ( n ) approaches infinity:

[ \lim_{n \to \infty} \frac{\frac{\sqrt{n}}{n}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{\sqrt{n} \cdot \sqrt{n}}{n} = \lim_{n \to \infty} 1 = 1 ]

Since the limit is a finite positive value, and since ( \sum \frac{1}{\sqrt{n}} ) converges, by the limit comparison test, ( \sum \frac{\sqrt{n}}{n} ) also converges.