how to find number of maximum 
Define $a_n=\frac{1000^n}{n!}$.Then $(a_n)_{n\geq1}$
(1) does not have a maximum
(2) attains maximum at exactly two values of $n$
(3) attains maximum at exactly one value of $n$
(4) attains maximum for infinitely many values of $n$

Trial: I take $\log$ and try to proceed.But I am stuck.Any help will be very helpful.Please help.

Note that $$\frac{a_{n+1}}{a_n} =\frac{1000^{n+1}}{(n+1)!}\cdot \frac{n!}{1000^n} =\frac{1000}{n+1}$$
 Now $\frac{a_{n+1}}{a_n} > 1$ for $n < 999$ again $\frac{a_{n+1}}{a_n} = 1$ for $n = 999$ and  $\frac{a_{n+1}}{a_n} < 1$ for $n > 999$
 So we get 
 $$a_1 < a_2 < a_3 < \cdots < a _{998} < a_{999} = a_{1000} > a_{1001} > \cdots$$ 
Clearly your option $(2)$ is correct.
