# `sum_(n=1)^oo 1/(n+3)` 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.

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`sum_(n=1)^oo1/(n+3)`

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/(n+3)`

Consider `f(x)=1/(x+3)`

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

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

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

`f'(x)=-1/(x+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 corresponding improper integral converges or diverges,

`int_1^oo1/(x+3)dx=lim_(b->oo)int_1^b1/(x+3)dx`

`=lim_(b->oo)[ln|x+3|]_1^b`

`=lim_(b->oo)ln|b+3|-ln|1+3|`

`=lim_(b->oo)ln|b+3|-ln4`

`lim_(b->oo)(b+3)=oo`

Apply the common limit:`lim_(u->oo)(ln(u))=oo`

`=oo-ln4`

`=oo`

Since the integral `int_1^oo1/(x+3)dx` diverges, we can conclude from the integral test that the series diverges.