`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.

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