sum_(n=1)^oo (n+2)/(n+1) 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)^oo(n+2)/(n+1)
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=(n+2)/(n+1)
Consider f(x)=(x+2)/(x+1)
Refer to the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for x>=1
We can also determine whether the function is decreasing by finding its derivative f'(x).
Apply quotient rule to find f'(x),
f'(x)=((x+1)d/dx(x+2)-(x+2)d/dx(x+1))/(x+1)^2
f'(x)=((x+1)-(x+2))/(x+1)^2
f'(x)=(x+1-x-2)/(x+1)^2
f'(x)=-1/(x+1)^2
f'(x)<0 which implies that f(x) is decreasing for x>=1
We can apply integral test, as the function satisfies all the conditions for the integral test.
Now let's determine whether the corresponding improper integral int_1^oo(x+2)/(x+1)dx converges or diverges,
int_1^oo(x+2)/(x+1)dx=lim_(b->oo)int_1^b(x+2)/(x+1)dx
Let's first evaluate the indefinite integral,
int(x+2)/(x+1)dx=int(x+1+1)/(x+1)dx
=int(1+1/(x+1))dx
Apply the sum rule,
=int1dx+int1/(x+1)dx
=x+ln|x+1|+C
int_1^oo(x+2)/(x+1)dx=lim_(b->oo)[x+ln|x+1|]_1^b
=lim_(b->oo)[b+ln|b+1|]-(1+ln|1+1|)
=lim_(b->oo)[b+ln|b+1|]-(1+ln2)
=oo-(1+ln2)
=oo
Since the integral int_1^oo(x+2)/(x+1)dx diverges, we conclude from the integral test that the series diverges.