## graydarl 3 years ago lin when n goes to infinity of ((2^n)*n!)/((n+1)*(n+2*....*(n+n)) Help me pleaseeeee

1. Zarkon

use the same trick I told you to use before..show that the sum of your sequence converges...therefore the limit is zero

2. graydarl

3. Zarkon

use the ratio test you could also use stirling's approximation to find the limt

4. graydarl

i write them in that form an+1 over an and i get stuck sorry i really don t know how to solve it, please give me a further hint or show me to see how u did it

5. Zarkon

$a_{n}=\frac{2^{n}{n}!}{\displaystyle\prod_{i=1}^{n}(n+i)}$ $a_{n+1}=\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}$ $\frac{a_{n+1}}{a_{n}}=\frac{\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}}{\frac{2^{n}{n}!}{\displaystyle\prod_{i=1}^{n}(n+i)}}$ $=\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\frac{\displaystyle\prod_{i=1}^{n}(n+i)}{2^{n}{n}!}$ $=2(n+1)\frac{\displaystyle\prod_{i=1}^{n}(n+i)}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}$ $=2(n+1)\frac{(n+1)(n+2)\cdots(n+n)}{(n+2)(n+3)+\cdots(n+n)(n+n+1)(n+n+2)}$ $=\frac{2(n+1)(n+1)}{(n+n+1)(n+n+2)}$ $=\frac{2(n+1)(n+1)}{(2n+1)(2n+2)}$ $=\frac{2(n+1)(n+1)}{(2n+1)2(n+1)}=\frac{n+1}{2n+1}\to\frac{1}{2} \text { as }n\to\infty$

6. Zarkon

$\frac{1}{2}<1$ so the sum converges and therefore the limit of the sequence is zero

7. Zarkon

do you understand?

8. graydarl

Yes, thank you very much, u hellped a lot :D