SOLUTION: For any integer n > 1, nth term = 1000^n / n! . Then the sequence {nth term} (A) does not have a maximum (B) attains maximum at exactly one value of n (C) attains maximum at ex

Question 854334: For any integer n > 1, nth term = 1000^n / n! . Then the sequence {nth term}
(A) does not have a maximum
(B) attains maximum at exactly one value of n
(C) attains maximum at exactly two values of n
(d) attains maximum for infinitely many values of n
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!
For any integer n > 1, nth term = 1000^n / n! . Then the sequence {nth term}
(A) does not have a maximum
(B) attains maximum at exactly one value of n
(C) attains maximum at exactly two values of n
(d) attains maximum for infinitely many values of n

The first term is 1000, the second term is 500000, so the sequence starts
out increasing.  Let's investigate to see for how many terms the sequence
is strictly increasing.

It will increase as long as a term is greater that the preceding term.

So we set the nth term greater than the n-1st term:



Multiply both sides by the LCD = n!



Divide both sides by 



So the 999th term is the last term for which the sequence increases.

The 999th term is 



Then the 1000th term is



which is equal to the 999th term since the last factor is  is 1.

After the 1000th term we will be multiplying by a fraction less than 1, 
beginning with multiplying by  and so the sequence will
begin to decrease with the 1001st term.

So the 999th and the 1000th terms, which are equal, are the largest
terms of the sequence.  Therefore

the sequence attains maximum at exactly two values of n, 999 and 1000.

Answer: (C)

Edwin

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