Using the integral test to test for divergence/convergence

[miglo]

Homework Statement

[tex]\sum_{n=1}^{\infty}\frac{8\arctan{n}}{1+n^2}[/tex]

Homework Equations

The Attempt at a Solution

so I am comparing it to the integral [itex]\int_{1}^{\infty}\frac{8\arctan{x}}{1+x^2}[/itex]
but at first i need to show that the function I am integrating is continuous, positive and decreasing. I know its continuous and positive from 1 to infinity but i need to show that it is decreasing
so i found the derivative of the function and got [itex]\frac{8-16x\arctan{x}}{(1+x^2)^2}[/itex] but i got stuck trying to find the critical points
specifically i forgot how to solve equations like [itex]8-16x\arctan{x}=0[/itex], mainly just because of that extra x in front of the arctan
i know this boils down to more of a precalc problem but i posted this on the calc homework help because i thought maybe some of my earlier steps were wrong

 

[Mark44]

Homework Statement

[tex]\sum_{n=1}^{\infty}\frac{8\arctan{n}}{1+n^2}[/tex]

Homework Equations

The Attempt at a Solution

so I am comparing it to the integral [itex]\int_{1}^{\infty}\frac{8\arctan{x}}{1+x^2}[/itex]
but at first i need to show that the function I am integrating is continuous, positive and decreasing. I know its continuous and positive from 1 to infinity but i need to show that it is decreasing
so i found the derivative of the function and got [itex]\frac{8-16x\arctan{x}}{(1+x^2)^2}[/itex] but i got stuck trying to find the critical points
specifically i forgot how to solve equations like [itex]8-16x\arctan{x}=0[/itex], mainly just because of that extra x in front of the arctan
i know this boils down to more of a precalc problem but i posted this on the calc homework help because i thought maybe some of my earlier steps were wrong

I don't see anything wrong with your work.

The equation you're trying to solve can be simplified to 2x arctan(x) = 1. I don't know of any ways to get analytic solutions to this equation, but approximation techniques yield solutions at [itex]x \approx \pm 0.765379[/itex], according to WolframAlpha. From the graph here, you can say that 1 - 2xarctan(x) < 0 for all x > 1, hence the derivative you've found is negative for x > 1.

 

[miglo]

yeah i tried wolfram alpha, but it didnt show any steps on how to solve that equation, was looking for an analytic solution but this will have to do
thanks mark44