Convergence of sequence given by $x_1=1$ and $x_{n+1}=x_n+\sqrt{x_n^2+1}$

Solution 1:

Hint3:

$$ \tan\frac{\theta}{2} = \frac{\sin\theta}{\cos\theta+1} $$ so $$ \cot\frac{\theta}{2} = \frac{\cos\theta+1}{\sin\theta} = \cot\theta+\csc\theta \\ =\cot\theta+\sqrt{\cot^2\theta+1} $$ for $0<\theta<\pi$, so that $\csc\theta>0$.

solution
Prove by induction $$ x_n = \cot\left(2^{-n}\frac{\pi}{2}\right) $$ so that $$ y_n = 2^n \tan\left(2^{-n}\frac{\pi}{2}\right) $$ which converges to $\pi/2$.

reference
Mathematics Magazine, Problem 1214:
solution published vol. 59, no. 2, April, 1986, p. 117

remark
This calculation can be viewed as:
The perimeter of a regular $2^n$-gon circumscribed about a circle converges to the circumference of the circle.