Question

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

Answer 1

The series:

`sum_(n=0)^oo sqrt(n+1)-sqrt(n)`

is divergent.

Explanation:

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

`s_n = sum_(k=0)^n = (cancel1-0) + (cancel(sqrt2)-cancel1) + (sqrt3-cancel(sqrt2))+...+(cancel(sqrt(n))-sqrt(n-1)) + (sqrt(n+1) -cancel(sqrt(n))) = sqrt(n+1)`

so that:

`lim_(n->oo) s_n = lim_(n->oo) sqrt(n+1) = oo`

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

`sqrt(n+1)-sqrt(n) = (sqrt(n+1)-sqrt(n))(sqrt(n+1)+sqrt(n))/(sqrt(n+1)+sqrt(n))`

Using the identity:

`(a+b)(a-b) = (a^2-b^2)`

this becomes:

`sqrt(n+1)-sqrt(n) = (canceln+1-canceln)/(sqrt(n+1)+sqrt(n)) = 1/(sqrt(n+1)+sqrt(n))`

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

`lim_(n->oo) (1/(sqrt(n+1)+sqrt(n)))/(1/n) = lim_(n->oo) n/(sqrt(n+1)+sqrt(n)) = oo`

which proves that:

`sum_(n=0)^oo sqrt(n+1)-sqrt(n) = oo`