Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ 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
(B). Take the $n$th root of $a _ { n }$ and $b _ { n }$ and use A.M. $\geq$ G.M.
Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ 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