Does $a_n = \cos\left(n\ln \left(1+\frac{\pi}{n}\right)\right)$ converge?

Question

I want to check if a sequence converges or diverges. The sequence is the following:

$$a_n = \cos\left(n\ln \left(1+\frac{\pi}{n}\right)\right)$$

I though of maybe using sandwich theorem, but can I use it, saying that the value will lie between $-1$ and $1$? If not, can anyone point me to where I should be looking at.

Answer 1

$$\begin{align}a_n &= \cos \left(n \ln \left( 1+ \frac \pi n\right) \right)\\ &=\cos\left(\ln\left(1+\frac\pi n\right)^n\right) \end{align}$$

You may recognize that $$\lim_{n\to \infty}\left(1+\frac x n\right)^n = e^x$$ for all $x$ (this being one way to define the exponential function).

Thus $\lim_{n\to\infty}\left(1+\frac \pi n\right)^n = e^\pi$.

Since $\ln$ is continuous throughout its domain, which includes $e^\pi$, and $\cos$ is continuous everywhere, your sequence will approach $\cos(\ln e^\pi)=\cos \pi = -1$.

Thanks for the tip on continuity from davin.

Answer 2

$$ a_n = \cos\left(n\ln \left(1+\frac{\pi}{n}\right)\right) = \cos\ln \left(\left(1+\frac 1{n/\pi}\right)^{n /\pi}\right)^\pi \to \cos\ln e^\pi = -1 $$

Answer 3

Note that near the origin $$\log(1+x)=x-\frac{x^2}{2}+O(x^2)$$

so that $$ n\log \left(1+\frac{\pi}{n}\right)={\pi}-\frac{\pi^2}{2n}+O\left(\frac{1}{n}\right)$$

It follows that the inner sequence goes to $\pi$ as $n\to\infty$ so the whole thing goes to $\cos \pi=-1$