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Let $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}, n\geq 1$, then $S_n^{\frac{1}{n}}$ converges to

  1. $e/2$
  2. $e/4$
  3. $e/8$
  4. $0$

Clearly $S_n^{\frac{1}{n}}=\frac{n}{(n+1)^{\frac{1}{n}}(n+2)^{\frac{1}{n}}...(n+n)^{\frac{1}{n}}}$, then how can we proceed

miracle173
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Riaz
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4 Answers4

8

We have $\frac{S_{n+1}}{S_n}=(1+\frac{1}{n})^n \frac{(n+1)^2}{(2n+1)(2n+2)} \to e/4.$

Hence $\lim S_n^{1/n}= \lim \frac{S_{n+1}}{S_n}=e/4.$

Fred
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3

You can rewrite $S_n$ as $S_n = \frac{n^n \cdot n!}{(2n)!}$ and then use Stirling's approximation: $$n!\sim\sqrt{2\pi n} (\frac{n}{e})^n$$ $$(2n)! \sim \sqrt{4\pi n} (\frac{2n}{e})^{2n}$$ Substituting these you get that: $$S_n \sim \frac{n^n \cdot \sqrt{2\pi n} (\frac{n}{e})^n}{\sqrt{4\pi n} (\frac{2n}{e})^{2n}} = \frac{e^n}{\sqrt2 \cdot 2^{2n}}$$

Taking the $n^{th}$ root, you get that $$S_n^{\frac{1}{n}} \rightarrow \frac{e}{4}$$

tia
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  • So if $a_n \sim b_n$ one always has $\lim\limits_{n\to\infty} a_n^{\frac{1}{n}} = \lim\limits_{n\to\infty} b_n^{\frac{1}{n}}$? Or when does this hold? – miracle173 Jun 06 '19 at 06:58
  • In general no, but Stirling's formula has very strong convergence properties, so I tend to treat it as an equality. If I were to use this in a paper I would prove it using inequalities. In particular, I would use the stronger version of Stirling's formula that states $\sqrt{2 \pi n}(\frac{n}{e})^n \leq n! \leq \sqrt{2 \pi n}(\frac{n}{e})^n e^{\frac{1}{12n}}$. – tia Jun 06 '19 at 07:19
  • @miracle173 Yes, if $a_n,b_n>0$ and $a_n\sim b_n$ then $a_n^{1/n}\sim b_n^{1/n}.$ – zhw. Jun 08 '19 at 16:19
2

We have $$(n+1)^n \lt (n+1)(n+2)...(n+n) \lt(2n)^n$$ and therefore $$\frac12=\frac{n}{2n}\le S_n^{\frac{1}{n}}\le \frac{n}{n+1}\le 1$$ and further $$\frac12 \le \lim\limits_{n\to 1}S_n^{\frac{1}{n}}\le 1$$ if that limit exists.

Only one of the proposed solution satisfies these inequalities. From $e=2.7\ldots$ follows $2.7<e<2.8$ and therefore

  1. $\frac{2.7}2 < \frac e 2 < \frac{2.8}{2}\implies 1.35 <\frac e2<1.4$
  2. $\frac{2.7}4 < \frac e 4 < \frac{2.8}{4}\implies 0.675 <\frac e4<0.7$
  3. $\frac{2.7}8 < \frac e 8 < \frac{2.8}{8}\implies 0.3375 <\frac e8<0.35$
  4. $0$

So if this limit exists and if it is one of these four numbers then it must be $\frac e 4$.

miracle173
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1

Another approach: $$\ln(S_n^{1/n})=-\frac1n\sum_{k=1}^n\ln\left(1+\frac kn\right)$$ which is a Riemann sum for the integral $$-\int_0^1\ln(1+x)\,dx.$$ So $$\lim_{n\to\infty}S_n^{1/n}=\exp\left(-\int_0^1\ln(1+x)\,dx\right)$$ so you just have to evaluate this integral.

Angina Seng
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