Using the integral test to test for divergence/convergence

miglo
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Homework Statement


\sum_{n=1}^{\infty}\frac{8\arctan{n}}{1+n^2}


Homework Equations





The Attempt at a Solution


so I am comparing it to the integral \int_{1}^{\infty}\frac{8\arctan{x}}{1+x^2}
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 \frac{8-16x\arctan{x}}{(1+x^2)^2} but i got stuck trying to find the critical points
specifically i forgot how to solve equations like 8-16x\arctan{x}=0, 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
 
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miglo said:

Homework Statement


\sum_{n=1}^{\infty}\frac{8\arctan{n}}{1+n^2}


Homework Equations





The Attempt at a Solution


so I am comparing it to the integral \int_{1}^{\infty}\frac{8\arctan{x}}{1+x^2}
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 \frac{8-16x\arctan{x}}{(1+x^2)^2} but i got stuck trying to find the critical points
specifically i forgot how to solve equations like 8-16x\arctan{x}=0, 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 x \approx \pm 0.765379, 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.
 
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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