Question

can some one do this
limit n->infinity (3^n -1)/n!

Answer 1

so taking the ratio of two succesive terms:
[(3*3^n - 1)/(n + 1)(n!)]*[n!/(3^n -1)] with a little algebraic manipulation you can see that this function is monotonically(sp?) decreasing. And by plugging in values, you can see that 0 < (3^n - 1)/(n!) < 2, so it must converge to a finite limit. Since it's always decreasing, it must approach zero. So lim n -> infty (3^n - 1)/(n!) = 0.

( i apologize for the messyness of this explanation, but i hope it's sufficient)

Answer 2

not that it really matters...is it \[\frac{3^n-1}{n!}\] or \[\frac{3^{n-1}}{n!}\]

Answer 3

it is the first

Answer 4

well...look at \[\frac{3^n}{n!}\]

\[=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdots 3}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdots n}\le \frac{3}{n}\to 0\text{ as }n\to\infty\]

Answer 5

@Zarkon This makes perfect sense, thank you very much.