`sum_(n=1)^oo 1/sqrt(n+2)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

Expert Answers

An illustration of the letter 'A' in a speech bubbles


The integral test is applicable if f is positive , continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.

For the given series `a_n=1/sqrt(n+2)`

Consider `f(x)=1/sqrt(x+2)`

Refer to the attached graph of the function. From the graph we observe that the function is positive and continuous for `x>=1`

Let's determine whether the function is decreasing by finding the derivative `f'(x)`




`f'(x)<0` which implies that the function is decreasing.

We can apply the integral test,since the function satisfies the conditions for the integral test.

Now let's determine whether the improper integral `int_1^oo1/sqrt(x+2)dx` converges or diverges.


Let's first evaluate the indefinite integral `int1/sqrt(x+2)dx`

Apply integral substitution:`u=x+2`



Apply the power rule,




Substitute back `u=x+2`

`=2sqrt(x+2)+C`  where C is a constant







Since the integral `int_1^oo1/sqrt(x+2)dx` diverges, we conclude from the integral test that the series also diverges.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Image (1 of 1)
Approved by eNotes Editorial