I want to prove that $\sum e^{-nx + \cos(nx)}$ is defined and continuous on the given interval of $(a, \infty)$ where $a >0$.
Then, how exactly do I show it is defined? It just seems trivial, since the exponential function, the linear function and cosine function are all "Defined" on the reals.
Further, to show it is continuous, I proceed as follows:
$$\sum e^{-nx + \cos(nx)}$$
Then, since each of the functions $f_n$ are continuous, by the algebraic continuity theorem, the partial sums will also be continuous, and so our sum is continuous. $$(\text{CAVEAT})$$ Before I can make the above bolded claim, I must first show the series is uniformly convergent, correct??
So, if that is the case, we use the M-test. We must find an $M_n > 0$ such that:
$$|f_n(x)| \leq M_n$$ holds.
$$|f_n(x)| = |e^{-nx + \cos(nx)}| = e^{-nx}e^{\cos(nx)}$$ since $\max \cos(nx) = 1$ $$\implies e^{-nx}e^{\cos(nx)} \leq e^{-nx}e$$
We look for these properties on the interval $(a, \infty)$
Thus
$$|f_n(x)| \leq e^{-ax}e = M_n$$
How can we conclude that $\sum M_n$ converges, however? Is this a trivial fact, since the exponential decreases so fast? I question this last step, because $\frac{1}{x} \to 0$ but the sum does not converge.
Once I have established continuity, how can I establish it is differentiable with a continuous derivative?
Any help is very much appreciated!
For the uniform convergence we have $$\sum_{n\geq1}e^{-nx+\cos\left(nx\right)}\leq\sum_{n\geq1}e^{-nx}\leq\sum_{n\geq1}e^{-na}=\frac{1}{e^{a}-1} $$ then we have uniform convergence by M-test. For the differentiation, we have to prove that $$f'\left(x\right)=\sum_{n\geq1}f_{n}'\left(x\right) $$ and so the uniform convergence of $\sum_{n\geq1}f_{n}'\left(x\right) $. Note that we have $$\left|\sum_{n\geq1}e^{-nx+\cos\left(nx\right)}\left(-n-n\sin\left(nx\right)\right)\right|\leq2\sum_{n\geq1}ne^{-na}=\frac{2e^{a}}{\left(e^{a}-1\right)^{2}} $$ then your claim.