Given sequences, finding the relation

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The discussion centers on comparing two sequences defined as $a_n=(1^2+2^2+ . . . +n^2)^n$ and $b_n=n^n(n!)^2$. The consensus is that $a_n < b_n$ for all $n > 1$, confirmed through the application of the AM-GM inequality. Initial attempts involved rewriting $a_n$ and comparing specific values, but the AM-GM approach proved to be more efficient. The conclusion is that the inequalities can be established definitively using this mathematical tool.

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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!
 
Last edited:
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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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