Show that the series $\sum 3^n \sin(\frac{1}{4^n} x)$ converges absolutely and uniformly on $(a,\infty)$, where $a>0$. Show that the series $\sum 3^n \sin(\frac{1}{4^n} x)$ converges absolutely and uniformly on $(a,\infty)$, where $a&gt;0$.

I am unable to determine how to apply the Weierstrass M-test here. Please show me how should I decide $M_n$ s.t. $|f_n(x)|\le M_n$.
$\sin x$ always less than $x$ for $x&gt;0$
$\implies \sin(1/4^n x) &lt; 1/4^n x$
$3^n \sin(1/4^nx) &lt; (3/4)^n \cdot 1/x$
$\max{ (3/4)^n \cdot 1/x}  =  (3/4)^n \cdot 1/a$  which converges
Hence by Weierestrass theory the given series converges uniformly.
You serie is differentiable


$|f_n(x)|\leq \dfrac{3^n}{4^n}x$ so it converges pointwise absolutely
$f_n$ is differentiable for all $n$
$|f_n|=\dfrac{3^n}{4^n}|\cos\left(\dfrac{x}{4^n}\right)|\leq\dfrac{3^n}{4^n}\triangleq M_n'$


$$ f'(x)=\sum \dfrac{3^n}{4^n}\cos\left(\dfrac{x}{4^n}\right)$$

By the derivation under sum sign theorem $f'$ exists and to answer your question

$\sum f_n $is uniformaly convergent also.

Note : I made a detour by the derivated serie to seek uniform convergence for the initial serie.