Question

How do you use the integral test to determine if `Sigma1/sqrt(n+1)` from `[1,oo)` is convergent or divergent?

Answer 1

The series is divergent

Explanation:

Apply the integral test only if the function `f(x)` is continuous, positive and decreasing.

The function `f(x)` is continuous as `AA x in [1,oo)` as `sqrt(1+x)` is continuous.

The function is positive as `sqrt(1+x)` is positive in the interval `[1,oo]`

`f'(x)=(1/sqrt(1+x))'=-1/2(1+x)^(-3/2)`

Therefore,

`f'(x)<0, AA x in [1,oo)`

So, we can apply the integral test.

Calculate the improper integral

`int_1^oodx/sqrt(1+x)`

First compute the indefinite integral

`intdx/sqrt(1+x)=int(1+x)^(-1/2)dx`

`=(1+x)^(-1/2+1)/(-1/2+1)`

`=2(1+x)^(1/2)`

Therefore,

`lim_(t->oo)int_1^tdx/sqrt(1+x)=lim_(t->oo)[2(1+x)^(1/2)]_1^t`

`=lim_(t->oo)(2sqrt(1+t)-2sqrt2)`

`=+oo`

As the improper integral is divergent , so the `sum_(n=1)^oo1/sqrt(1+n)` is divergent