Nice hope so

Think about the sum of a geometric sequence (e^ix/3)^m. Taking a real part of this sum, we get a result.

We need to find a pattern for $\cos mx$ , which can be given by the Chebyshev polynomials . On this helpful page we find that

$\cos mx = T_m(\cos x)$

as well as

$\sum_{m=0}^\infty T_m(z)t^m = \frac{1-tz}{1-2tz+t^2}.$

We substitute $t=z=\frac13$ :

$\sum_{m=0}^\infty \frac{\cos mx}{3^m}=\frac{1-\frac13\cdot\frac13}{1-2\cdot\frac13\cdot\frac13+\left(\frac13\right)^2}=\boxed{1}.$