# calculating limit of a sequence

• December 30th 2012, 10:56 AM
Stormey
calculating limit of a sequence
i need to calculate the limit of this sequence:

$\sqrt[n^2]{n!}$

i tried to use the sandwich rule to squeeze it between to sequences, but got stuck.

this is what i did so far:

$\sqrt[n^2]{n!}<\sqrt[n^2]{n\cdot n\cdot \cdot \cdot \cdot \cdot n}=\sqrt[n^2]{n^n}=\sqrt[n]{n}\rightarrow 1$

but i couldn't find anything smaller then the given sequence that can be easily calculated.
for example, i know that:

$\sqrt[n^2]{n}<\sqrt[n^2]{n!}$

and i know that $\sqrt[n]{n}\rightarrow 1$, but is it enough to say that if $\sqrt[n]{n}\rightarrow 1$ then also $\sqrt[n^2]{n}\rightarrow 1$?

BTW, if there's any other way to get to the answer ( $\sqrt[n^2]{n!}\rightarrow 1$) i'd love to hear.
$\sqrt[n^2]{n}=e^{(\ln n)/n^2}\to1$ because $(\ln n)/n^2\to0$ as $n\to\infty$. In fact, without using $n=e^{\ln n}$, the function $a^{1/x}$ where a > 1 monotonically decreases as x increases. Therefore, $n^{1/n^2}. On the other hand, $n^{1/n^2}>1$ for n > 1. Therefore, the fact that $n^{1/n}\to1$ indeed implies that $n^{1/n^2}\to1$.