So, about the following limit:
$$\lim_{n\to\infty} \left( \frac{1+\cos(\frac{1}{2^{n}})}{2} \right)^n$$
I tried several things to evaluate it, namely looking at it as $\cos(\frac{1}{2^{n+1}})^{2n}$ instead or as $\exp(2n \cdot \ln({\cos(\frac{1}{2^{n+1}})})$ and then trying to show that the limit of $n\cdot\ln({\cos(\frac{1}{2^{n+1}})})$ is $0$ (for example using L'Hopital's rule), but I haven't been very successful (though it is possible I gave up too early). I'm just not sure how to approach this cosine-exponential combo. I believe the limit is $1$, so things like the root test weren't very helpful in this case, either.
I'd really appreciate a direction/hint, a full solution, or anything inbetween.
Best Answer
\begin{align*} \left(\dfrac{1+\cos(1/2^{n})}{2}\right)^{n}\leq\left(\dfrac{1+1}{2}\right)^{n}=1. \end{align*} Now $\cos u\geq 1-u^{2}/2$ for small $u\geq 0$, then \begin{align*} \left(\dfrac{1+\cos(1/2^{n})}{2}\right)^{n}\geq\left(1-\dfrac{1}{2^{2(n+1)}}\right)^{n}\rightarrow 1. \end{align*}