Sol040506

$\mbox{Let }\theta_n = \arctan(n)\mbox{. Prove that, for } n = 1,2,3,\cdots\,$

$\theta_{n+1}-\theta_n < \frac{1}{n^2+n}$

Solution I by landen

$\mbox{Define: }\phi_n=\frac{\pi}{2}-\theta_n\,$

$\tan\phi_n=\frac{1}{n}\mbox{ by trig identity}\,$

$\mbox{Then by subtraction: }\theta_{n+1}-\theta_n = \arctan\frac{1}{n}-\arctan\frac{1}{n+1}$

$\mbox{Expanding the arctans in a Taylor series:}\,$

$\theta_{n+1}-\theta_n = \left(\frac{1}{n}-\frac{1}{n+1}\right) - \frac{1}{3}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\cdots$

Since the series is alternating and the terms decrease in absolute value, we can truncate after one term with a negative error:

$\theta_{n+1}-\theta_n < \left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{n^2+n}\qquad\square$

Solution II by landen

$\mbox{From a Taylor series for }x\in[0,\pi/2);\quad \tan(x)=x+x^3/3+2x^5/15\cdots\ >x$

$\tan(\theta_{n+1}-\theta_n) = \frac{\tan\theta_{n+1}-\tan\theta_n}{1+\tan\theta_{n+1}\,\tan\theta_n}\mbox{ famous trig identity}$

$\theta_{n+1}-\theta_n < \tan(\theta_{n+1}-\theta_n) = \frac{\tan\theta_{n+1}-\tan\theta_n}{1+\tan\theta_{n+1}\,\tan\theta_n}=\frac{n+1-n}{1+n(n+1)}<\frac{1}{n^2+n}\qquad \square$

Solution III by Bumperoni

$\mbox{Consider the continuous function }f(x)=\arctan(x)+1/x\,$

$\mbox{We can prove equivalently: }f(n+1)-f(n)<0\,$ ;

$f'(x)=\frac{1}{x^2+1}-\frac{1}{x^2}<0$

$f'(c)=\frac{f(n+1)-f(n)}{n+1-n}\mbox{ with }c\in[n,n+1]\mbox{ by mean value theorem}$

$f(n+1)-f(n)=f'(c)<0\qquad \square$

Retrieved from "http://www.efnet-math.org/w/Sol040506"