How to prove $\lim_{n\to{\infty}}\frac{x^n}{n!} = 0$
My try:
I know that factorial function increases much more rapidly than exponential function but how to really prove it.
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1I think you want the limit at $\infty$. Consider what happens when $n>|x|$. – David Mitra Feb 23 '14 at 10:42
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1I think you can use the $n$-th power test to show that the series $\sum_n \frac{x^n}{n!} (=: e^x)$ converges. Then, its terms will of course go to zero :-) – Singhal Feb 23 '14 at 10:45
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Are you allowed to use Stirling's approximation? – Hoseyn Heydari Feb 23 '14 at 10:49
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@HoseynHeydari Never heard of it and not in course also. – user2369284 Feb 23 '14 at 10:50
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@DavidMitra we expand the numerator as (xxx...n times) and denominator as (123...n). Now we segregate the fraction as (x(x/2)(x/3)*...(x/n)). As x/n will tend to 0 so the whole fraction will tend to 0. Is this correct? – user2369284 Feb 23 '14 at 11:00
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Take an $n_0 \in \mathbb N$ such that $n_0 > |x|$. Then $$\left|\frac{x^n}{n!}\right| = \frac {|x|}1\cdot\frac {|x|}2\dotsm\frac {|x|}{n_0}\dotsm\frac {|x|}n$$ For all $n \ge n_0$ we have $\frac {|x|}n \le \frac {|x|}{n_0} < 1$. That means $$\left|\frac{x^n}{n!}\right| \le \frac{{|x|}^{n_0-1}}{(n_0-1)!} \cdot \left(\frac {|x|}{n_0}\right)^{n-n_0}$$ which is a geometric sequence which converges to zero since $\frac{|x|}{n_0} < 1$.
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I had already corrected the mistake much before you gave the answer – user2369284 Feb 23 '14 at 11:01