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Prove that for real number $p\geq 0$ we have

$$\lim_{n\to\infty}\frac{(1^{1^p} \cdot2^{2^p}\cdots n^{n^p})^\frac{1}{n^{p+1}}}{n^\frac{1}{(p+1)}} = \exp\left(\dfrac{-1}{(p+1)^2}\right).$$

metamorphy
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user10024395
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  • Start by taking the natural log of the expression under the limit and try to view it as a Riemann sum corresponding to the integral: $\displaystyle \int_{0}^{1} x^p \ln{x}dx$. – hot_queen Apr 02 '14 at 04:29
  • Is there any other way to do it without using integral. My lecturer does not allow integral. – user10024395 Apr 02 '14 at 12:10

2 Answers2

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As several answers/hints have pointed out: consider the natural log of the LHS. We get $$\frac{1^p\ln(1)+2^p\ln(2)+..+n^p\ln(n)}{n^{p+1}}-\frac{\ln(n)}{p+1}=\frac{(p+1)1^p\ln(1)+..+(p+1)(n)^p\ln(n)-n^{p+1}\ln(n)}{(p+1)n^{p+1}}$$

Then we apply the theorem of Stolz-Cesaro, as

\begin{align}\lim_{n\to\infty}&\frac{(p+1)(n+1)^p\ln(n)-(n+1)^{p+1}\ln(n+1)+n^{p+1}\ln(n)}{(p+1)((p+1)n^p+...+1)}\\&=\lim_{n\to\infty}\frac{(p+1)n^p\ln(n)-n^{p+1}\ln(n+1)-(p+1)n^p\ln(n)+n^{p+1}\ln(n)+o(n^p)}{(p+1)((p+1)n^p+o(n^p))}\\&=\lim_{n\to\infty}\frac{n\ln\left(1-\frac{1}{n+1}\right)+\frac{o(n^p)}{n^p}}{(p+1)((p+1)+\frac{o(n^p)}{n^p})}\\&=\frac{-1}{(p+1)^2}\end{align}

The various $o(n^p)$ are sum of the terms of the form $n^k\ln(n+1)$ or $n^k$, $k<n$ in the numerator and denominator respectively. It is clear why these terms are equal to $o(n^p)$

A Nonny
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  • @metamorphy thanks for leading to this solution :) But I don't understand why $\ln(n)$ is written instead of $\ln(n+1)$ in some places while using Stolz-Cesaro... It would be greatly appreciated :) – VIVID Apr 12 '20 at 14:36
  • I think the author has been inactive for some time. – VIVID Apr 12 '20 at 14:37
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A couple thoughts here:
1) L'Hopital's rule might be helpful. At the very least, it is worth exploring.
2) The advise to consider the log of the expression seems like a good idea, even if you do not use the integral to sum up the result. If the limit of the natural log of the l.h.s expression is $-1/(p+1)^2$ then you have your result.

Bill Province
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