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help me please true or fulse

(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1? $$ \begin{array}{l} u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\quad ;u_n > 0 \\ \Leftrightarrow \ln \left( {u_n } \right) = \ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right)^{\frac{1}{n}} \\ \Leftrightarrow \ln \left( {u_n } \right) = \frac{1}{n}\ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right) \\ \Leftrightarrow \ln \left( {u_n } \right) = \frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n} \\ \Leftrightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\ln \left( {u_n } \right)} \right) = \mathop {\lim }\limits_{n \to + \infty } \left[ {\frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n}} \right] = 0 \\ \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1 \\ \end{array} (2)

$$ \begin{array}{l} t_n = \frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }} \\ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{1 \times 3 \times \cdots (2n - 1)(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{{1 \times 3 \times \cdots (2n - 1)}} \\ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{1} \\ = \mathop {\lim }\limits_{n \to + \infty } \left( {2\left( {\frac{n}{{n + 1}}} \right)^{n + 1} + \frac{{n^n }}{{\left( {n + 1} \right)^{n + 1} }}} \right) = 2e^{ - 1} ;\quad \left( {\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{x}{n}} \right)^n = e^x } \right) \\ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{\frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }}}} = 2e^{ - 1} \\ \end{array} $$

(3) $$\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$$

$$\left( 4 \right)\mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = !? $$

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You can rewrite the definition of $v_n$ as $$ v_1=2\\[2ex] v_{n+1}=\frac{(2n+1)(2n+2)}{n}v_n $$ so $$ \frac{v_{n+1}}{v_n}=\frac{(2n+1)(2n+2)}{n} $$

Another way to see this is noting that $$ v_n=\frac{(2n)!}{(n-1)!} $$

For $t_n$ you have \begin{align} t_n &=\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{n^n}\\ &=\frac{1}{n^n} \frac{(1\cdot3\cdot\ldots\cdot(2n-1))(2\cdot 4\cdot\ldots\cdot 2n)} {2\cdot 4\cdot\ldots\cdot 2n}\\ &=\frac{1}{n^n}\frac{(2n)!}{2^n\cdot n!}\\ \end{align} so $$ \frac{t_{n+1}}{t_n}= \frac{1}{(n+1)^{n+1}}\frac{(2n+2)!}{2^{n+1}\cdot (n+1)!} n^n\frac{2^n\cdot n!}{(2n)!} =\frac{1}{2}\frac{n^n}{(n+1)^{n+1}}\frac{(2n+2)(2n+1)}{n+1} $$ Thus we need to look at $$ \lim_{n\to\infty}\frac{n^n}{(n+1)^{n+1}}(2n+1) $$ and it's easier doing the limit of the inverse: $$ \lim_{n\to\infty}\frac{(n+1)^{n+1}}{n^n}\frac{1}{2n+1}= \lim_{n\to\infty}\frac{(n+1)^n}{n^n}\frac{n+1}{2n+1}= \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\frac{n+1}{2n+1}=\frac{e}{2} $$ so your given limit is $2/e$.

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  • $\begingroup$ I have need to be clarified for the first limite $\endgroup$ – user194150 Nov 22 '14 at 13:07
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    $\begingroup$ @user194150 What in particular? The limit is $\infty$. $\endgroup$ – egreg Nov 22 '14 at 14:57
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$$\underset{n\rightarrow\infty}{\lim}\frac{v_{n+1}}{v_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1\right)\cdots\left(n+1+n-1\right)\left(n+1+n\right)\left(n+1+n+1\right)}{n\cdots\left(n+n\right)}=$$ $$=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1+n\right)\left(n+1+n+1\right)}{n}=\infty$$ because you have essentialy $n^{2}$ at numerator.$$\underset{n\rightarrow\infty}{\lim}\frac{t_{n+1}}{t_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{1\cdot3\cdots\left(2n-1\right)\left(2n+1\right)}{\left(n+1\right)^{n+1}}\cdot\frac{n^{n}}{1\cdot3\cdots\left(2n-1\right)}=$$ $$=\underset{n\rightarrow\infty}{\lim}\left(2n+1\right)\frac{n^{n}}{\left(n+1\right)^{n+1}}=\underset{n\rightarrow\infty}{\lim}\left(2\left(\frac{n}{n+1}\right)^{n+1}+\frac{n^{n}}{\left(n+1\right)^{n+1}}\right)=2e^{-1}$$ because $\underset{n\rightarrow\infty}{\lim}\left(1+\frac{x}{n}\right)^{n}=e^{x}.$

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$$

\begin{array}{l} n! = \sqrt {n\pi e} \left( {\frac{n}{e}} \right)^n \\ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\sqrt {\left( {2n} \right)\pi e} \left( {\frac{{2n}}{e}} \right)^{2n} }}{{\sqrt {\left( {3n} \right)\pi e} \left( {\frac{{3n}}{e}} \right)^{3n} }}}}} \right) \\ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{4}{{27}}e\sqrt[n]{{\sqrt {\frac{2}{3}} }}} \right) = \frac{4}{{27}}e\quad \\ ;\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt[n]{{\sqrt {\frac{2}{3}} }}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{2}{3}} \right)^{\frac{1}{{2n}}} = 1 \\ \end{array}

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