- #1
Saitama
- 4,243
- 93
Problem:
Define $a_n=(1^2+2^2+ . . . +n^2)^n$ and $b_n=n^n(n!)^2$. Recall $n!$ is the product of the first n natural numbers. Then,
(A)$a_n < b_n$ for all $n > 1$
(B)$a_n > b_n$ for all $n > 1$
(C)$a_n = b_n$ for infinitely many n
(D)None of the above
Attempt:
The given sequence $a_n$ can be written as
$$a_n=\frac{n^n(n+1)^n(2n+1)^n}{6^n}$$
But I am not sure what to do now. I understand that this is a very less attempt towards the given problem but I really have no clue how someone should go about comparing these kind of sequences. Please give a few hints.
Any help is appreciated. Thanks!
Define $a_n=(1^2+2^2+ . . . +n^2)^n$ and $b_n=n^n(n!)^2$. Recall $n!$ is the product of the first n natural numbers. Then,
(A)$a_n < b_n$ for all $n > 1$
(B)$a_n > b_n$ for all $n > 1$
(C)$a_n = b_n$ for infinitely many n
(D)None of the above
Attempt:
The given sequence $a_n$ can be written as
$$a_n=\frac{n^n(n+1)^n(2n+1)^n}{6^n}$$
But I am not sure what to do now. I understand that this is a very less attempt towards the given problem but I really have no clue how someone should go about comparing these kind of sequences. Please give a few hints.
Any help is appreciated. Thanks!
Last edited: