<s>[INST] The decay of the series given by $x_n=x_{n-1}+\cos x_{n-1}$. Let $x_n=x_{n-1}+\cos x_{n-1}$, $x_1=1$. It is easy to see that $x_n\to \dfrac{\pi}{2}$. However, how can we show that $n^n(x_n-\dfrac{\pi}{2})\to 0\ (n\to\infty)$?
I find Stolz formula hard to use.
 [/INST] We can manipulate the recursion as follows:
\begin{align}
x_n &= x_{n-1}+\cos x_{n-1}
\\
\dfrac{\pi}{2} - x_n &= \dfrac{\pi}{2} - x_{n-1} - \cos x_{n-1}
\\
\dfrac{\pi}{2} - x_n &= \dfrac{\pi}{2} - x_{n-1} - \sin\left(\dfrac{\pi}{2}-x_{n-1}\right)
\end{align}
Since $|\theta-\sin \theta| \le \dfrac{1}{6}|\theta|^3$ for all $\theta \in \mathbb{R}$, we have that $$\left|\dfrac{\pi}{2}-x_n\right| \le \dfrac{1}{6}\left|\dfrac{\pi}{2}-x_{n-1}\right|^3$$ for all integers $n \ge 2$. We can then use induction to show that $$\left|\dfrac{\pi}{2}-x_n\right| \le \sqrt{6}\left(\dfrac{\pi/2-1}{\sqrt{6}}\right)^{3^{n-1}}.$$
Note that $\dfrac{\pi/2-1}{\sqrt{6}} \approx 0.233 < 1$. From here, it is easy to see that $$\ln\left[n^n\left|\dfrac{\pi}{2}-x_n\right|\right] \le 3^{n-1} \cdot \ln\left(\dfrac{\pi/2-1}{\sqrt{6}}\right)+n\ln n - \dfrac{1}{2}\ln 6 \to -\infty,$$ and thus, $n^n\left|\dfrac{\pi}{2}-x_n\right| \to 0$ as $n \to \infty$.
</s>