Question

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

Answer 1

The series:

`sum_(n=1)^oo (sqrt(n+2)-sqrt(n))/n`

is convergent.

Explanation:

Rationalize the numerator of `a_n`:

`a_n = (sqrt(n+2)-sqrt(n))/n = ((sqrt(n+2)-sqrt(n))/n)((sqrt(n+2)+sqrt(n))/(sqrt(n+2)+sqrt(n))) = (n+2-n)/(n(sqrt(n+2)+sqrt(n))) = 2/(n(sqrt(n+2)+sqrt(n))`

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

`a_n = 2/(n(sqrt(n+2)+sqrt(n))) < 2/(n(sqrt(n)+sqrt(n))) = 2/(2nsqrtn) =1/n^(3/2)`

We know that the series:

`sum_(n=1)^oo 1/n^(3/2)`

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

`sum_(n=1)^oo (sqrt(n+2)-sqrt(n))/n`

is also convergent.