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anonymous
 3 years ago
I must show that lim whn n goes to infinity of (1/n!)*((n/3)^n)=0
anonymous
 3 years ago
I must show that lim whn n goes to infinity of (1/n!)*((n/3)^n)=0

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n! }\left( \frac{ n }{ 3 } \right)^n=0\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0One slick way is to show that \[\sum_{n=1}^{\infty}\frac{ 1 }{ n! }\left( \frac{ n }{ 3 } \right)^n\] converges using the ratio test if the series is finite then the limit of the sequence is zero

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am still stuck, i dont t know how to solve it or if what i try is correct :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know how to apply the ratio test that Zarkon was talking about? I just took a go at it (was good memory jog) and I had to realise that I didn't really care about knowing the actual limit of the thing you're interested in when applying the ratio test, I only cared about making sure it was less than 1 in the limit.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i tryed something and i am not sure that is correct so i can t tell that i know how to apply it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well you want to prove that series Zarkon mentioned is convergent. Have a read of this: http://en.wikipedia.org/wiki/Ratio_test Using the terminology from that page, your a_n is 1/n! * (n/3)^n. Now to apply the ratio test, you want to see if a_{n+1} / a_n converges to something strictly less than 1. Take the thing we defined as a_n, and then have a look at what a_{n+1} / a_n looks like. Some things will cancel. Then see if that thing converges to something strictly less than 1. You should be able to see that it does (it's bounded and monotonic). So you've applied the ratio test and found that the result is that the series Zarkon described does converge. A series converges only if the elements in the series converge to zero in the limit. That gives you the result you wanted.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.youtube.com/watch?v=iy8mhbZTY7g http://www.youtube.com/watch?v=AwJ0P8B25tY  First video is a tutorial on how to use the ratio test and what it means. Second one explains how to deal with factorials using the ratio test.
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