# `sum_(n=1)^oo arctan(n)/(n^2+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.

## Expert Answers

`sum_(n=1)^ooarctan(n)/(n^2+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=arctan(n)/(n^2+1)`

Consider `f(x)=arctan(x)/(x^2+1)`

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

We an 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^ooarctan(x)/(x^2+1)dx` converges or diverges.

`int_1^ooarctan(x)/(x^2+1)dx=lim_(b->oo)int_1^b arctan(x)/(x^2+1)dx`

Let's first evaluate the indefinite integral `intarctan(x)/(x^2+1)dx` ,

Apply integral substitution:`u=arctan(x)`

`du=1/(x^2+1)dx`

`=intudu`

Apply power rule,

`=u^2/2`

Substitute back `u=arctan(x)`

`=1/2(arctan(x))^2+C` where C is a constant

`int_1^ooarctan(x)/(x^2+1)dx=lim_(b->oo)[1/2(arctan(x))^2]_1^b`

`=lim_(b->oo)1/2[(arctan(b))^2-(arctan(1))^2]`

`=1/2[(pi/2)^2-(pi/4)^2]`

`=1/2(pi^2/4-pi^2/16)`

`=1/2((4pi^2-pi^2)/16)`

`=3/32pi^2`

Since the integral `int_1^ooarctan(x)/(x^2+1)dx` converges, we conclude from the integral test that the series converges.

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