1. Convergene of series

I think I copied this down right. It's a problem our professor gave us at the end of class, and his handwriting is a little sketchy. Also, I think I'm suppose to test to see if it converges.

problem:

(1/3)+(1*4)/ (3*5)+(1*4*7)/ (3*5*7)+..+[1*4*7*..*(3n-2)] /[3*5*7*..*(2n+1)]

Can I get a hint? Just a shot in the dark here, but it looks recursive.. kinda like the fibonacci sequence (the future term depends on the previous one); but this is a series. So.. not too sure what to do.

 Sorry about the typo in the title! Totally meant "convergence".

2. Originally Posted by Hikari
I think I copied this down right. It's a problem our professor gave us at the end of class, and his handwriting is a little sketchy. Also, I think I'm suppose to test to see if it converges.

problem:

(1/3)+(1*4)/ (3*5)+(1*4*7)/ (3*5*7)+..+[1*4*7*..*(3n-2)] /[3*5*7*..*(2n+1)]

Can I just get a hint?
Just a hint.

For n > 3, (3n-2) > (2n+1)

3. Are you suggesting a comparison test?

Could I compare that whole crazy series like this:

crazy series > [SUM] (3n-2)/(2n+1)

And then use a test for divergence on the smaller one? And then by comparison test to say that because the smaller one diverges, so does the bigger one? Sorry, I'm rather tired, so if that made absolutely no sense, I apologize.

4. was I wrong?

5. $\displaystyle s_n = \sum_{n=0}^{\infty} \frac{3n-2}{2n+1}$

The divergence test says: if the $\displaystyle \lim_{n-\infty} a_n$ does not equal 0 or it does not exist, then the series $\displaystyle s_n$ diverges.

$\displaystyle \lim_{n-\infty} \frac{3n-2}{2n+1} = \frac{3}{2}$, using L'Hospital rule (sp?). Therefore, the series diverges.