Hi
I was wondering if we would use the Ratio Test on this series? If not any ideas?
Sigma: 2.4.6.....(2n) divided by 3.5.7.....(2n+1)
$\displaystyle a_{n}=\frac{2\cdots(2n)}{3\cdots(2n+1)}$ and so $\displaystyle \frac{a_{n+1}}{a_n}=\frac{2\cdots(2n)\cdot(2n+2)}{ 3\cdots\cdot(2n+1)\cdot(2n+3)}\cdot\frac{3\cdots(2 n+1)}{2\cdots(2n)}=\frac{2n+2}{2n+3}$. So the ratio test doesn't work. But...try showing that it's always greater than something....any guess what?
Use the inequality
$\displaystyle \frac{1}{n+1} < \frac{2\cdots(2n)}{3\cdots(2n+1)}$
then do direct comparison with the harmonic series.
Edit: you can prove the inequality by induction on n.
$\displaystyle \frac{1}{2} < \frac{2}{3}$ proves it true for n=1.
Now assume it is true for $\displaystyle k \leq n$.
Then
$\displaystyle \frac{1}{n+1} < \frac{2\cdots(2n)}{3\cdots(2n+1)}$
$\displaystyle \frac{1}{n+1} \cdot \frac{2n+2}{2n+3} < \frac{2\cdots(2n)}{3\cdots(2n+1)} \cdot \frac{2n+2}{2n+3}$
$\displaystyle \frac{1}{n+1} \cdot \frac{n+1}{n+2} < \frac{1}{n+1} \cdot \frac{2n+2}{2n+3} < \frac{2\cdots(2n)}{3\cdots(2n+1)} \cdot \frac{2n+2}{2n+3}$
$\displaystyle \frac{1}{n+2} < \frac{2\cdots(2n+2)}{3\cdots(2n+3)}$
and the induction is complete.