show that the following series are absolutely convergent Please help - I need to show that the following series are absolutely convergent.
     a)      $\displaystyle\sum_{n=2}^{\infty}\frac{\sin(n\theta)}{2^n}$
     b)
     $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}{\sin(\pi/n)}$
     c)
     $\displaystyle\sum_{n=2}^{\infty}\frac{\cos(n\theta)}{\sqrt{n^3}-1}$
 A: Hint #1:
For a) b) and c) take the comparism test.
Hint #2:
For b) use that $\sin(\frac{\pi}{n})\leq \frac{\pi}{n}$
A: Hints:
a) The absolute value of the general term is $\le \dfrac{1}{2^n}$.  And it is a standard fact that $\displaystyle\sum_{n=2}^\infty \dfrac{1}{2^n}$ converges. It is a geometric series, you can even find its sum explicitly.  So by Comparison, the series we started with converges absolutely.
b) Let's look at the absolute value of the general term. The $(-1)^n$ part disappears. It is a general fact that $|\sin x|\le |x|$.
So the absolute value of the general term is $\le \dfrac{1}{\sqrt{n}}\cdot\dfrac{\pi}{n}$. This is $\dfrac{\pi}{n^{3/2}}$.
It is a standard fact that $\displaystyle\sum_1^\infty \dfrac{1}{n^p}$ converges if $p\gt 1$. Now use Comparison.
c) The absolute value of the general term is $\le \dfrac{1}{\sqrt{n^3-1}}$. Note that if $n\ge 2$, then $n^3-1 \ge n^3/8$. (This is a deliberately sloppy inequality. we could write down a sharper one.)
It follows that $\dfrac{1}{\sqrt{n^3-1}}\le \dfrac{2}{n^{3/2}}$. Now do a comparison much like that in b). 
