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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

  1. $(x_n)$ converges but $(y_n)$ does not converge
  2. $(y_n)$ converges but $(x_n)$ does not converge
  3. both $(x_n)$ and $(y_n)$ converge
  4. 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.

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The answer is 1. $x_n$ converges but not $y_n$. Try solving it now with this info. If you get stuck, let us know. –  Gautam Shenoy Nov 19 '12 at 14:23
    
If I can show that (yn) converges and since we notice that xn<yn ,then by comparison test we can say that (xn) also converges. But i can not show that (yn) converges. I want to mention that i have proved that (xn) converges to 1. Any hint in this regard will be helpful. –  learner Nov 19 '12 at 14:24
    
@Gautam thanks.then option (3) will be correct,i guess. –  learner Nov 19 '12 at 14:26
    
@GautamShenoy thanks for the hint. –  learner Nov 19 '12 at 14:47
    
Sterling's approximation. Search on wikipedia –  Adam Rubinson Nov 19 '12 at 15:08
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3 Answers

up vote 1 down vote accepted

$\log(y_n)=\frac{\log(n!)}{n}$. Rewrite this to $\frac{\log(1)+\log(2)+\ldots +\log(n)}{1+1+\ldots + 1}$. This has the limit $\infty$. Then, so does $y_n$.

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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 $(\sqrt[n]{a_n})$.

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you can use the stirling formula http://www.sosmath.com/calculus/sequence/stirling/stirling.html

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