1. ## Series- Real Analysis

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)

2. Sorry but i dont think anyone will understand that. use /frac{}{} and \sum between tags to make it clearer.

3. Originally Posted by hebby
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?

4. The Raabe Test, could be used here and also we can compare it with a harmonic series 1/2n+1, which diverges and thus the given series diverges?

5. 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.

6. can we also use the Raabe test?...an+1/an => 1-a/n

7. Originally Posted by hebby
can we also use the Raabe test?...an+1/an => 1-a/n
I am unfamiliar with that test.

8. Originally Posted by hebby
can we also use the Raabe test?...an+1/an => 1-a/n
Yes.

9. Wow how could you you figure that out!