sequence of functions-convergence

Question

So i need help analyzing convergence of this sequence:

$$(f_{n}), f_{n}:[-\pi,\pi] \rightarrow \mathbb{R}, f_{n}=\frac{n (cosx)^{2n-1}}{4n+1}$$ for n $\in \mathbb{N}$

Does it converge pointwise? Does it converge uniformly?

Any help would be appreciated. Thank you in advance.

Answer 1

The limit is easily calculated $$f_{\infty}(x)=\left\{\begin{matrix}1/4 & \cos(x)=1 \\ -1/4 & \cos(x)=-1 \\ 0 & \text{otherwise}\end{matrix}\right.$$ which is discontinuous. This should tell you something about whether or not the sequence converges uniformly.

Answer 2

Since $-1<\cos (x)<1$ then so is $(\cos(x))^{2n-1}$, this means that $$\lim_{n\rightarrow \infty}\frac{n\cos(x)^{2n-1}}{4n+1}=\lim_{n\rightarrow \infty}\frac{\cos(x)^{2n-1}}{4}$$

This limit has several values depending on the value of $x$, i.e. $f_{\infty}$ is discontinuous, and this case arises when the convergence is not uniform.