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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.

Deepak Suwalka
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xldd
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    Well, $x+\cos x$ has zero derivative at fixed point, which kind of guarantees quadratic convergence of iterations (or better). Basically, $|x_n-{\pi\over2}|\sim\exp(-\exp(n))$, with some coefficients and stuff. – Ivan Neretin Dec 27 '16 at 06:46
  • @IvanNeretin How can we deduce $|x_n-{\pi\over2}|\sim\exp(-\exp(n))$?? Is there any references to locate in? – xldd Dec 27 '16 at 07:10

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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 $n \to \infty$.

JimmyK4542
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