Finding the limit for recurrence relation $ x_{n+1} = \sqrt{x_n + \frac 14} - \frac 1 2 $ Let $(x_n)_{n \ge1}$ be a sequence defined by $x_0 > 0$ and $x_{n+1}$ = $\sqrt{x_n + \frac 14} - \frac 12$. Show that $\lim_{n\to\infty}x_n = 0$ and find $\lim_{n\to\infty}nx_n$.
I'm guessing I'd first have to show that the sequence is monotonically decreasing and bounded, thus convergent, in which case, finding the actual limit would not be that difficult, yet the proper way to do that seems to elude me.  Also, for the second task I tried applying Stolz-Cesaro which only led me to a dead end. Simply put, I'm stuck. Any hint would be greatly appreciated.
Thanks.

$$
0 < x_{n+1} = \sqrt{x_n + \frac 14} - \frac 1 2 = \frac{x_n}{\sqrt{x_n + \frac 14} + \frac 1 2} < x_n
$$
shows that the sequence is decreasing and bounded below by zero, therefore convergent. The limit $L$ must satisfy
$$
 L = \sqrt{L + \frac 14} - \frac 1 2
$$
which implies that $L= 0$.
Then Stolz-Cesaro shows that
$$
 \lim_{n \to \infty}\frac{1}{n x_n} = \lim_{n \to \infty}\left(\frac{1}{x_{n+1}} - \frac{1}{x_n} \right)
$$
provided that the latter limit exists. But
$$
 \frac{1}{x_{n+1}} - \frac{1}{x_n} = \frac{1}{\sqrt{x_n + \frac 14} - \frac 1 2} - \frac{1}{x_n} = \frac{1}{\sqrt{x_n + \frac 14} + \frac 1 2}
$$
converges to one since $x_n$ converges to zero.
