MHB Given sequences, finding the relation

AI Thread Summary
The discussion revolves around comparing the sequences defined as \( a_n = (1^2 + 2^2 + \ldots + n^2)^n \) and \( b_n = n^n(n!)^2 \). The initial approach suggests rewriting \( a_n \) for easier comparison, but the user expresses uncertainty about the next steps. A suggestion is made to compare specific values, such as \( a_2 \) and \( b_2 \), to eliminate options regarding their relationship. Another participant proposes using the AM-GM inequality as a quicker method to determine the relationship between the sequences. Ultimately, the discussion highlights the complexity of comparing these sequences and the utility of mathematical inequalities in solving such problems.
Saitama
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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!
 
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Start by comparing $$a_2$$ and $b_2$ this will eliminate one of the first two inequalities . Then proceed by induction.
 
ZaidAlyafey said:
Start by comparing $$a_2$$ and $b_2$ this will eliminate one of the first two inequalities . Then proceed by induction.

Thanks ZaidAlyafey but I seem to have figured out a better solution. Use of AM-GM inequality gives the answer in a few seconds. :)
 
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