## graydarl Group Title lin when n goes to infinity of ((2^n)*n!)/((n+1)*(n+2*....*(n+n)) Help me pleaseeeee one year ago one year ago

1. Zarkon Group Title

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 Group Title

3. Zarkon Group Title

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

4. graydarl Group Title

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 Group Title

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

6. Zarkon Group Title

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

7. Zarkon Group Title

do you understand?

8. graydarl Group Title

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