Be $(x_n)_n\ge 1 $ such that $x_1=1$ and $x_{n+1}=x_n+\sqrt{x_n^2+1}$ for every $n\ge 1$. Prove that the sequence $y_n=(2^n/x_n)_n \ge 1$ is convergent and find it s limit.
Being positive and decreasing, $y_n$ is clearly convergent. But finding it's limit really put me intro trouble. Any help? Thank you!
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.
