# calculating limit of a sequence

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

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

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

this is what i did so far:

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

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

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

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

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