# Homework Help: Finding the general term of the series

1. Sep 21, 2014

### smart_worker

1. The problem statement, all variables and given/known data

Find the general term and test the nature(convergent/divergent) of:
1/3 + 2/15 + 2/35 + .....

3. The attempt at a solution
If I simplify I get ,
1/3 + (1)(2)/(3)(5) + (1)(2)(3)/(3)(5)(7) +.............

I found that,
TERM(N) = TERM(N-1)((N)/(2N+1)

After this I am struck.

2. Sep 21, 2014

### smart_worker

The new format of this website is difficult for me.

3. Sep 21, 2014

### HallsofIvy

You see, I suspect, that the numerator of the nth term is n! The denominator is something like a factorial except that it involves only odd integers- okay, make it a factorial by inserting the even integers:
$$\frac{1}{3*5*7}= \frac{2*4*6}{2*3*4*5*6*7}= \frac{2*4*6}{7!}$$

Now, factor a 2 out of each term in the numerator: 2*4*6= 2(1)*2(2)*2*3= 2^3(3!).

So $$\frac{1}{3*5*7}= \frac{2*4*6}{2*3*4*5*6*7}= \frac{2^3(3!)}{7!}$$

Your term $$\frac{1*2*3}{3*5*7}= \frac{2^3(3!)^2}{7!}$$.

Can you write the nth term now?

Do you see n

4. Sep 21, 2014

### smart_worker

yeah... i can write the nth term and i found that the series is convergent.

But this series:
1/4 + (1)(5)/(4)(8) + (1)(5)(9)/(4)(8)(12) + ...

The general term for the denominator is 4^n (n!)

but the numerator terms are tricky : 1*5*9

==> 1*2*3*4*5*6*7*8*9/2*3*4*6*7*8

==>9!/2*3*4*6*7*8

how do i simplify this part : 2*3*4*6*7*8 ?