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Compute $$\lim\limits_{n\to\infty} \frac {n}{\sqrt[n]{n!}} $$ and $$\lim\limits_{n\to\infty} {2n \choose n}$$

I suppose that the first limit is equal to $e $.

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Hint for the first limit: $$ \ln \left( \lim\limits_{n\to\infty} \frac {n}{\sqrt[n]{n!}} \right) = \lim\limits_{n\to\infty} \left( \ln \frac {n}{\sqrt[n]{n!}} \right) = \lim\limits_{n\to\infty} \left( \ln n- \frac{1}{n}\sum_{j=1}^n \ln j \right). $$

And I suppose there is something wrong with your second limit since it is obviously divergent.

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First Limit

Using Riemann-Sums, we get $$ \begin{align} \lim_{n\to\infty}\frac1n\log\left(\frac{n!}{n^n}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac kn\right)\frac1n\\ &=\int_0^1\log(x)\,\mathrm{d}x\\[9pt] &=-1 \end{align} $$ Thus, $$ \lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}=\frac1e $$ and therefore, $$ \lim_{n\to\infty}\frac{n}{(n!)^{1/n}}=e $$


Second Limit

Hint: $$ \binom{2n}{n}=\binom{2n-2}{n-1}\left(4-\frac2n\right) $$

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From Taylor expansion we have $\frac{x^n}{n!} \leq e^x$ for any $x ≥ 0$. Spe- cialising this to $x = n$ we obtain a crude lower bound

$n! ≥ n^ne^{−n}$.

In the other direction, we trivially have

$n! ≤ n^n$

so we know already that n! is within an exponential factor of $n^n$. One can do better by starting with the identity

$log n! = \sum_{m=1}^n log m$

and viewing the right-hand side as a Riemann integral approximation to $\int_1^nlog x\ dx$. Indeed a simple area comparison yields the inequalities

$\int_1^nlog x dx ≤\sum_{m=1}^n log m ≤ log n +\int_1^n log x dx$

which leads to the inequalities $en^ne^{−n} ≤ n! ≤ en × n^ne^{−n}$

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  • $\begingroup$ Do you mean $\frac{x^n}{n!} \leq e^x$? $\endgroup$ – robjohn Oct 28 '17 at 7:55
  • $\begingroup$ Corrected, thanks. $\endgroup$ – Hu xiyu Oct 28 '17 at 7:56
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Let $a_n=\frac{n^n}{n!}.$ Thus, $$\lim_{n\rightarrow+\infty}\frac{n}{\sqrt[n]{n!}}=\lim_{n\rightarrow+\infty}\sqrt[n]{a_n}=\lim_{n\rightarrow+\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n=e.$$

The second: $$\binom{2n}{n}=\frac{2n\cdot(2n-1)\cdot...\cdot(n+1)}{n\cdot(n-1)\cdot...\cdot1}\geq2\cdot2\cdot...\cdot2=2^n\rightarrow+\infty.$$

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