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graydarl
lin when n goes to infinity of ((2^n)*n!)/((n+1)*(n+2*....*(n+n)) Help me pleaseeeee
use the same trick I told you to use before..show that the sum of your sequence converges...therefore the limit is zero
i can t figure it out please explain further please
use the ratio test you could also use stirling's approximation to find the limt
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
\[a_{n}=\frac{2^{n}{n}!}{\displaystyle\prod_{i=1}^{n}(n+i)}\] \[a_{n+1}=\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\] \[\frac{a_{n+1}}{a_{n}}=\frac{\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}}{\frac{2^{n}{n}!}{\displaystyle\prod_{i=1}^{n}(n+i)}}\] \[=\frac{2^{n+1}{(n+1)}!}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\frac{\displaystyle\prod_{i=1}^{n}(n+i)}{2^{n}{n}!}\] \[=2(n+1)\frac{\displaystyle\prod_{i=1}^{n}(n+i)}{\displaystyle\prod_{i=1}^{n+1}(n+1+i)}\] \[=2(n+1)\frac{(n+1)(n+2)\cdots(n+n)}{(n+2)(n+3)+\cdots(n+n)(n+n+1)(n+n+2)}\] \[=\frac{2(n+1)(n+1)}{(n+n+1)(n+n+2)}\] \[=\frac{2(n+1)(n+1)}{(2n+1)(2n+2)}\] \[=\frac{2(n+1)(n+1)}{(2n+1)2(n+1)}=\frac{n+1}{2n+1}\to\frac{1}{2} \text { as }n\to\infty\]
\[\frac{1}{2}<1\] so the sum converges and therefore the limit of the sequence is zero
Yes, thank you very much, u hellped a lot :D