# Thread: Limits / Series question

1. ## Limits / 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.

2. Originally Posted by champrock
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.
Factorise $\frac{n^3-1}{n^3+1}$ as $\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$, and notice that $n^2+n+1 = (n+1)^2 - (n+1) + 1$. You will then find that you have a telescoping product.

3. 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

4. You easily undestand if you write the sequence of terms...

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

... 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...

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

Very nice!...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by champrock
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
Use the hint I gave before: $n^2+n+1 = (n+1)^2 - (n+1) + 1$. That tells you that the term $n^2+n+1$ in the numerator of each fraction cancels with the term $(n+1)^2 - (n+1) + 1$ in the denominator of the next one.

6. I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3

7. The term of order n is...

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

For $n=3$ is...

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

Kind regards

$\chi$ $\sigma$