If \(p\) is a non-zero real number, and \(x_1,x_2,\dots,x_n\) are positive real numbers, then the power mean with exponent \(p\) of these positive real numbers is:

\[M_p(x_1,x_2,\dots,x_n)=\left(\frac 1n\sum_{i=1}^n x_i^p\right)^{\frac 1p}.\]

For \(p=0\) we set it equal to the geometric mean (which is the limit of means with exponents approaching zero):

\[M_0(x_1,x_2,\dots,x_n)=\sqrt[n]{\prod_{i=1}^n x_i}.\]

Evaluate the limit below for all real number \(p\):

\[\lim_{n\to\infty}\sqrt[n]{M_p\left(\binom n0,\binom n1,\dots,\binom nn\right)}.\]

I got the incomplete answer (see problems Interesting Limit 24, 21 and power mean inequality):

\[\lim_{n\to\infty}\sqrt[n]{M_p\left(\binom n0,\binom n1,\dots,\binom nn\right)}= \begin{cases} 1,&\text{if }p\le -1\\ \sqrt e,&\text{if }p=0\\ 2,&\text{if }p\ge 1 \end{cases}\]

Using Aaghaz Mahajan's result below, further answer is obtained:

\[\lim_{n\to\infty}\sqrt[n]{M_p\left(\binom n0,\binom n1,\dots,\binom nn\right)}= \begin{cases} 1,&\text{if }p\le -1\\ \sqrt e,&\text{if }p=0\\ 2,&\text{if }p> 0 \end{cases}\]

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## Comments

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TopNewest@Brian Lie Hello!! I think this result might help you for positive 'p'...........I have read it in a book, I can send you the book if you want.......

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Dear Aaghaz, my email address is lixuepeili@gmail.com. Thank you for your sending.

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@Brian Lie Sir, could you suggest me any books to learn Dynamics from?? Or Physics in general??

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Mechanics and Special Relativity by Freddie Williams is more mathematical.

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Thanks Sir....!!

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May I say something? I asked same question to Josh Silverman 3 years ago, he answered Kleppner and Klowenkow is the best.

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Ohhhh!!! Thanks for the information!! I'll download it right away..... :)

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