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 Post subject: hard limit (high school)
PostPosted: Fri, 19 Jun 2009 12:06:21 UTC 
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Joined: Fri, 24 Oct 2008 21:17:32 UTC
Posts: 31
Location: Cluj Napoca
Here is the limit

\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt[{{2^n}}]{{{n^{\left( {\begin{array}{*{20}{c}}
   n  \\
   0  \\
\end{array}} \right)}}{{(n + 1)}^{\left( {\begin{array}{*{20}{c}}
   n  \\
   1  \\
\end{array}} \right)}}...{{(2n)}^{\left( {\begin{array}{*{20}{c}}
   n  \\
   n  \\
\end{array}} \right)}}}}}}{n}

hint:

Spoiler:
\[{\text{Let }}f{\text{:[0,1]}} \to \mathbb{R}{\text{ a function such that }}\left| {f(x) - f(y)} \right| \leqslant \left| {x - y} \right|\]
\[\forall x,y \in {\text{[0,1] and }}{{\text{B}}_n}{\text{:[0,1]}} \to \mathbb{R}{\text{, }}{{\text{B}}_n}(x) = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}
   n  \\
   k  \\

 \end{array} } \right)} {x^k}{(1 - x)^{n - k}}f\left( {\frac{k}
{n}} \right)\]
\[{\text{with the convetion }}{0^0} = 1;{\text{ Prove that }}\left| {{{\text{B}}_n}(x) - f(x)} \right| \leqslant \frac{1}
{{2\sqrt n }}{\text{ }}\forall n \in {\mathbb{N}^*}\]
\[{\text{and  }}\forall x \in {\text{[0,1] using Cauchy - Schwartz inequality}}{\text{.}}\]


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