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.

sum_(n=1)^oo1/sqrt(n+2)
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)=(-1/2)(x+2)^(-1/2-1)
f'(x)=-1/2(x+2)^(-3/2)
f'(x)=-1/(2(x+2)^(3/2))
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.
int_1^oo1/sqrt(x+2)dx=lim_(b->oo)int_1^b1/sqrt(x+2)dx
Let's first evaluate the indefinite integral int1/sqrt(x+2)dx
Apply integral substitution:u=x+2
=>du=dx
int1/sqrt(x+2)dx=int1/sqrt(u)du
Apply the power rule,
=(u^(-1/2+1)/(-1/2+1))
=2u^(1/2)
=2sqrt(u)
Substitute back u=x+2
=2sqrt(x+2)+C  where C is a constant
int_1^oo1/sqrt(x+2)=lim_(b->oo)[2sqrt(x+2)]_1^b
=lim_(b->oo)[2sqrt(b+2)-2sqrt(1+2)]
=2lim_(b->oo)sqrt(b+2)-2sqrt(3)
=2(oo)-2sqrt(3)
=oo-2sqrt(3)
=oo
Since the integral int_1^oo1/sqrt(x+2)dx diverges, we conclude from the integral test that the series also diverges.