comparison between two sequences I stumble upon this problem which says:
Let $x_n=n^{1/n}$ and $y_n=(n!)^{1/n},\  n\ge1$ be 
two sequences of real numbers. Then


*

*$(x_n)$ converges but $(y_n)$ does not converge

*$(y_n)$ converges but $(x_n)$ does not converge

*both $(x_n)$ and $(y_n)$ converge

*neither $(x_n)$ nor $(y_n)$ converges


If I can show that $(y_n)$ converges and since 
we notice that $x_n<y_n$ ,then by comparison 
test we can say that $(x_n)$ also converges. But 
i can not show that $(y_n)$ converges. I want 
to mention that i have proved that $(x_n)$ converges to $1$.
 Any hint
in this regard will be helpful.
 A: $\log(y_n)=\frac{\log(n!)}{n}$. Rewrite this to $\frac{\log(1)+\log(2)+\ldots +\log(n)}{1+1+\ldots + 1}$. This has the limit $\infty$. Then, so does $y_n$. 
A: For any sequence $(a_n)$ of positive numbers, we have $$\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}\leq\liminf_{n\to\infty}\sqrt[n]{a_n}\leq\limsup_{n\to\infty}\sqrt[n]{a_n}\leq\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}.$$
The above chain of inequalities lets us show that if $\cfrac{a_{n+1}}{a_n}$ converges, or diverges to $\infty$, then (respectively) so does $\sqrt[n]{a_n}$--moreover, in the case of convergence, they share the same limit. We can use this approach to show that $(x_n)$ converges and that $(y_n)$ diverges to $\infty$.
Even if $\cfrac{a_{n+1}}{a_n}$ neither converges nor diverges to $\infty$--that is, if $\liminf\limits_{n\to\infty}\cfrac{a_{n+1}}{a_n}<\limsup\limits_{n\to\infty}\cfrac{a_{n+1}}{a_n}$--it is still possible that $\sqrt[n]{a_n}$ can converge or diverge to $\infty$, so this method doesn't work all the time. Still, it's handy to keep it in our toolbox when we're dealing with sequences of the form $(\sqrt[n]{a_n})$.
A: you can use the stirling formula http://www.sosmath.com/calculus/sequence/stirling/stirling.html 
