- Thread starter
- #1

i must calculate :

LIM (when n goes to infinit) ( (n^2 + n + 1) * ln( (n+1)/(n+2) ) * ln ( (2n+1)/(2n+3) ) )

i know i should use the case of 1 ^ infinit but i can't get it right .

thanks in advance for every answer

- Thread starter gambix
- Start date

- Thread starter
- #1

i must calculate :

LIM (when n goes to infinit) ( (n^2 + n + 1) * ln( (n+1)/(n+2) ) * ln ( (2n+1)/(2n+3) ) )

i know i should use the case of 1 ^ infinit but i can't get it right .

thanks in advance for every answer

- Admin
- #2

I would try L'Hôpital's Rule myself. Can you rewrite the expression so that the limit is of the indeterminate form \(\displaystyle \frac{0}{0}\)?

- Moderator
- #3

- Feb 7, 2012

- 2,701

I would start by writing the limit as \(\displaystyle \lim_{n\to\infty}\Bigl(1+ \frac1n + \frac1{n^2}\Bigr) \Bigl(n\ln\frac{n+1}{n+2}\Bigr) \Bigl(n\ln\frac{2n+1}{2n+3}\Bigr)\) (dividing the first factor by $n^2$ and multiplying each of the other factors by $n$). If you can show that the limit of each of those three factors is $1$ then you can use the theorem that the limit of a product is the product of the limits.

i must calculate :

LIM (when n goes to infinit) ( (n^2 + n + 1) * ln( (n+1)/(n+2) ) * ln ( (2n+1)/(2n+3) ) )

i know i should use the case of 1 ^ infinit but i can't get it right .

thanks in advance for every answer