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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
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

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ZarkonBest ResponseYou've already chosen the best response.3
use the same trick I told you to use before..show that the sum of your sequence converges...therefore the limit is zero
 one year ago

graydarlBest ResponseYou've already chosen the best response.0
i can t figure it out please explain further please
 one year ago

ZarkonBest ResponseYou've already chosen the best response.3
use the ratio test you could also use stirling's approximation to find the limt
 one year ago

graydarlBest ResponseYou've already chosen the best response.0
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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.3
\[a_{n}=\frac{2^{n}{n}!}{\displaystyle\prod_{i=1}^{n}(n+i)}\] \[a_{n+1}=\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\] \[\frac{a_{n+1}}{a_{n}}=\frac{\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}}{\frac{2^{n}{n}!}{\displaystyle\prod_{i=1}^{n}(n+i)}}\] \[=\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\frac{\displaystyle\prod_{i=1}^{n}(n+i)}{2^{n}{n}!}\] \[=2(n+1)\frac{\displaystyle\prod_{i=1}^{n}(n+i)}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\] \[=2(n+1)\frac{(n+1)(n+2)\cdots(n+n)}{(n+2)(n+3)+\cdots(n+n)(n+n+1)(n+n+2)}\] \[=\frac{2(n+1)(n+1)}{(n+n+1)(n+n+2)}\] \[=\frac{2(n+1)(n+1)}{(2n+1)(2n+2)}\] \[=\frac{2(n+1)(n+1)}{(2n+1)2(n+1)}=\frac{n+1}{2n+1}\to\frac{1}{2} \text { as }n\to\infty\]
 one year ago

ZarkonBest ResponseYou've already chosen the best response.3
\[\frac{1}{2}<1\] so the sum converges and therefore the limit of the sequence is zero
 one year ago

graydarlBest ResponseYou've already chosen the best response.0
Yes, thank you very much, u hellped a lot :D
 one year ago
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