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Mathematics
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lim(n^n/n!) as n-->(infinity). How to tackle this beast?
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Put it at the top of e's
\[ \huge e^{\ln \frac{n^n}{n!}} = \huge e^{\ln n^n - \ln n!} = e^{n \ln n - \ln n!}\] And \( \ln n! = n \ln n - n\) <--- a good approximation for lage n
ahhhhh. Thanks. Owe you guys some posts, i'll pay my dues soon. Thanks again.
yw!
Write \[ \frac { n^n}{n!} = \frac { n\ n\ n \cdots n \ n} { 1\ 2\ 3 \cdots (n-1)\ n }= n \frac n 2\frac n 3 \cdots \frac {n}{n-1} \frac n n > n \] So the limit is Infinity
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@experimentX, did you see my solution above. It is similar to yours so as to say that the ratio behave like something bigger than n. It does not use the approximation of n! for large n.
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