How do you test the series #Sigma (sqrt(n+2)-sqrtn)/n# from n is #[1,oo)# for convergence?

To test the series ( \sum_{n=1}^{\infty} \frac{\sqrt{n+2} - \sqrt{n}}{n} ) for convergence, we can use the Limit Comparison Test or the Ratio Test.

Using the Limit Comparison Test:

1. Choose another series ( \sum_{n=1}^{\infty} b_n ) that converges or diverges and is somewhat similar to the given series.
2. Take the limit as ( n ) approaches infinity of the ratio ( \frac{a_n}{b_n} ), where ( a_n = \frac{\sqrt{n+2} - \sqrt{n}}{n} ).

Using the Ratio Test:

1. Take the limit as ( n ) approaches infinity of the absolute value of the ratio of successive terms: ( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ), where ( a_n = \frac{\sqrt{n+2} - \sqrt{n}}{n} ).

If the limit obtained in either test is finite and nonzero, then the series diverges. If the limit is zero, then the test is inconclusive, and another test may need to be used. If the limit is one, then the test is also inconclusive, and other methods like the Integral Test or Comparison Test may be considered.

Now if we decrease the denominator we obtain a sequence that is greater:

We know that the series:

is convergent based on the p-series test, so the series:

is also convergent.