Determine whether the series $\sum_{n=1}^\infty\left({\arctan(n)-\frac{\pi}{2}}\right) $ converge

Question

I am trying to determine whether the series $$\sum_{n=1}^\infty \left({\arctan(n)-\frac{\pi}{2}}\right) $$ converges from my calculus homework.

Im not sure which candidate to choose for comprasion test.

By looking at the graphs of $f(x) = 1/x$ And the graph of $\arctan(x) - \pi/2$ It looks like $f(x)$ ~ $\arctan(x) - \pi/2$ but this is not formal ofcourse.

I will need some assistance please. Thanks!

Answer 1

You have\begin{align}\lim_{x\to\infty}\frac{\frac\pi2-\arctan(x)}{\frac1x}&=\lim_{x\to\infty}\frac{-\frac1{x^2+1}}{-\frac1{x^2}}\\&=\lim_{x\to\infty}\frac{x^2}{x^2+1}\\&=1.\end{align}So,$$\lim_{n\to\infty}\frac{\frac\pi2-\arctan(n)}{\frac1n}=1$$and therefore, since the harmonic series diverges, your series diverges.

Answer 2

Let $x_n=(\pi /2)-\arctan n.$ Then $\pi /4\ge x_n>0$ . And we have $\tan x_n=1/n.$ Therefore $$x_n>\sin x_n=(\cos x_n)/n\ge (\cos \pi/4)/n=1/(n\sqrt 2).$$