Thread: series help

1. series help

help i think that the n on the 2 is a misprint
still help me to please

2. Re: series help

The $2^n$ in the denominator? The series wont converge without it so it looks fine to me.

3. Re: series help

Originally Posted by bkbowser
The $2^n$ in the denominator? The series wont converge without it so it looks fine to me.
What do you get as 1st 3 terms of this series?
I'm a bit confused! Thanks.
Trying to set up a looper...

4. Re: series help

Originally Posted by DenisB
What do you get as 1st 3 terms of this series?
I'm a bit confused! Thanks.
Trying to set up a looper...
Denis here is the general term.
$\Large{{a_n} = \frac{{\prod\limits_{k = 1}^n {(2k - 1)} }}{{{2^n}\left[ {\prod\limits_{k = 1}^n {(3k - 2)} } \right]}}}$

5. Re: series help

How in hell did I write down 2^4 instead of 2^n !!
Thanks Plato....

6. Re: series help

how do you solve it??

7. Re: series help

$\displaystyle \Large{{a_n} = \frac{{\prod\limits_{k = 1}^n {(2k - 1)} }}{{{\prod\limits_{k = 1}^n {(3k - 2)} } }}} <1$

8. Re: series help

u and a = numerator calculator
v and b = denominator calculator

Looper program:

u=1:v=1:a=1:b=1
loop n from 1 to ?
a=a*u
b=b*v
t=a/(b*2^n) : current term value
s=s+t : sum so far
u=u+2
v=v+3
next n

s = ~.79025 at n=15 and basically stays at that for eternity!!
All kinda "silly" to me