Does the infinite series $\sum_{n=0}^{\infty}e^{-nx}\cos{nx}$ converge point-wise for all $x \in [1, \infty)$?
Facts that help: $$|\cos{nx}| \leq 1 \\ x \geq 1 \quad \forall \space te^{t} > 0 \\ e^{-nx} \leq e^{-n} \quad \forall \space x \geq 1$$
We need to find a series that is less than or equal to the series above. and that i can't do :(
I think the series we searching is 1/n^2 and the sum of it is 2. Is it right? edit: damn its from n=0 not from n=1 so I have no idea what to do :(
I think I have some ideas that could help you out.
Check out this link https://en.wikipedia.org/wiki/Weierstrass_M-test
With the Weierstrass test you prove something stronger than point-wise and that is the uniform convergence.
In a few words, if something converges in an interval uniformly then it must have converged pointwise. So... here I want to use the Weierstrass test in order to do that I must $$|e^{-nx}cos(nx)| \leq |e^{-nx}|\leq |e^{-n}|=e^{-n}=M_{n}$$ So all we need is prove that the following series converges $$\sum_{n=1}^\infty M_n$$ In order to prove that I used the Ratio Test https://en.wikipedia.org/wiki/Ratio_test
$$\lim_{x\to \infty}|\frac{M_{n+1}}{M_n}|=\lim_{x\to \infty}|\frac{e^{-(n+1)}}{e^{-n}}|=e^{-1}<1$$ So, as the series with Mn converges, we can use the Weierstrass test to say that: $$\sum_{n=1}^\infty e^{-nx}cos(nx)$$ converges absolutely and uniformly on x $\epsilon$ [1,$\infty$).
And as it converges uniformly on [1,$\infty$) in particular it converges point-wise on [1,$\infty$). I know that it starts with n=0 but as it converges uniformly with n=1 then it also converges uniformly with n=0, it's just adding 1 to the series.
I hope this helps