hi

Let

P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]

n=2,3,4,......

Find Limit P when N tends to infinity.

Printable View

- Apr 3rd 2009, 12:36 AMchamprockLimits / Series question
hi

Let

P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]

n=2,3,4,......

Find Limit P when N tends to infinity. - Apr 3rd 2009, 03:53 AMOpalg
- Apr 3rd 2009, 04:13 AMchamprock
what should i do with n^2 - n + 1 ?

I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1 - Apr 3rd 2009, 04:50 AMchisigma
You easily undestand if you write the sequence of terms...

$\displaystyle P= \prod_{n=2}^{\infty} \frac{(n-1)\cdot (n^{2} + n + 1)}{(n+1)\cdot (n^{2} - n + 1)}= \frac {1\cdot 7}{3\cdot 3}\cdot \frac {2\cdot 13}{4\cdot 7}\cdot \frac {3\cdot 21}{5\cdot 13}\cdot \frac {4\cdot 31}{6\cdot 21}\cdot \dots$

... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...]

Finally is...

$\displaystyle P=\frac{2}{3}$

Very nice!(Clapping)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 3rd 2009, 04:54 AMOpalg
- Apr 3rd 2009, 05:21 AMchamprock
http://img3.imageshack.us/img3/4053/43200964415pm.png

I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3 - Apr 3rd 2009, 05:49 AMchisigma
The term of order n is...

$\displaystyle a_{n}= \frac{(n-1)\cdot (n^{2} +n + 1)}{(n+1)\cdot (n^{2} - n + 1)}$

For $\displaystyle n=3$ is...

$\displaystyle a_{3}= \frac{2\cdot 13 }{4\cdot 7}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$