# Series with Trigonometric funtions

1. Nov 18, 2007

### azatkgz

[SOLVED] Series with Trigonometric funtions

1. The problem statement, all variables and given/known data
Determine whether the series converges conditionally,converges absolutely or diverges.

$$\sum_{n=2}^{\infty}\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}$$

3. The attempt at a solution

$$\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}=\frac{\sin n\cos \frac{1}{n}+\sin \frac{1}{n}\cos n}{\ln (\ln n)}=\frac{\sin n(1+O(\frac{1}{n^2}))+\cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}$$

$$\sum_{n=2}^{\infty}\sin n,\sum_{n=2}^{\infty}\cos n$$ are bounded

From Dirichlet's Test we can deduce that this series converges

$$|a_n|=\left|\frac{\sin n(1+O(\frac{1}{n^2})}{\ln (\ln n)}\right|+\left|\frac{cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}\right|$$

if we look for example at $$\left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (\ln n)}\right|>\left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (n)}\right|$$

0<|cosn|<1 and $$\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}$$ diverges by Integral Test
So this series converges conditionally!!

2. Nov 19, 2007

### Gib Z

Completely correct, nothing I can really add of value. You seem to be getting most of these right =]