Cauchy criterion approximation

Question

I have to use Cauchy's criterion on $$\sum_{n=1}^\infty \frac{\cos{n}a}{3^n}.$$ Because I'll have the absolute value of it to prove Cauchy and cos n is from [-1,1], so it will actually be from [0,1], can I approximate the fraction to just $\frac{a}{3^n}?$ Thanks!

Answer 1

Notice that $-1 \leq \cos(na) \leq 1$ for any $n$ and $a$. Also, $\sum_{n=i}^{j-1} \frac{1}{3^n} = \frac{3}{2}(3^{-i}-3^{-j})$ is an easy geometric series.