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

Question

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.

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

score 8 · Accepted Answer · answered Feb 22 '13 at 15:20

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.

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

1 Answers1