How do you test the series #Sigma sqrt(n+1)-sqrtn# from n is #[0,oo)# for convergence?

To test the series (\sum_{n=0}^{\infty} (\sqrt{n+1} - \sqrt{n})) for convergence, you can use the telescoping series test.

Here's how:

1. First, express each term as a difference of squares to simplify the series.

2. Observe the pattern of cancellation to determine if the series converges or diverges.

Let's proceed:

1. Express each term as a difference of squares: [\sqrt{n+1} - \sqrt{n} = (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{(\sqrt{n+1})^2 - (\sqrt{n})^2}{\sqrt{n+1} + \sqrt{n}} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}]

2. Now, observe the pattern of cancellation: [\frac{1}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \times \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}]

   Notice that each term cancels out with the next term, leaving only the first term (\sqrt{1} - \sqrt{0} = 1).

Since the series reduces to a finite value (1), the series (\sum_{n=0}^{\infty} (\sqrt{n+1} - \sqrt{n})) converges.

We can see that the series is divergent by analyzing the partial sums:

so that:

We can however use a convergence test in the following way: we have that:

Using the identity:

this becomes:

Now we can use the limit comparison test with the harmonic series:

which proves that: