# The largest term in the sequence $x_n=\frac {1000^n}{n!}$ for $n=1,2,3,…$

I faced the following problem which says:

The largest term in the sequence $x_n=\frac {1000^n}{n!}$ for $n=1,2,3,.......$ is which of the following ?
$1.$ is $x_{999}$
$2.$ is $x_{1001}$
$3.$ is $x_{1}$
$4.$ does not exist.

My Attempt: I see that $x_n-x_{n-1}=\frac {1000^{n-1}}{(n-1)!} \times ( \frac {1000}{n}-1).$ Now It is clear that $x_{n-1} < x_n$ if $n<1000.$ So $x_{998}<x_{999}=x_{1000}.$ But the sequence is a decreasing sequence for $n > 1000.$

I am not sure how to progress hereon or draw a conclusion from what I have found out. Can someone point me in the right direction?Thanks in advance for your time.

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So, it increases upto 999 stays the same at 1000 and decreases after that. So the biggest is...? – Ishan Banerjee Feb 22 '13 at 15:20
Do you know which of 1) and 4) above were deemed the correct answer? – DJohnM Feb 22 '13 at 15:48

It is at least as eay to consider $\frac{x_n}{x_{n-1}}$ and check when this quotient is $>1$ or $<1$. Note that $$\frac{x_n}{x_{n-1}}=\frac{1000}n,$$ hence (as you seem to also have found) $$x_1<x_2<\ldots <x_{999}=x_{1000} >x_{1001}> \ldots$$
Thus $x_{999}$ is a largest term, but the largest term does not exist.