I found this limit calculation problem in a book. For a real number $p\geq 0$ we have $$\lim_{n\rightarrow \infty}\frac{\left (1^{1^p}2^{2^p}\dots n^{n^p}\right )^{\frac{1}{n^{p+1}}}}{n^{\frac{1}{p+1}}}=e^{-\frac{1}{(p+1)^2}}$$

After taking logarithm this is equivalent to showing $$\sum_{k=1}^n \frac{1}{n}{\left ( \frac{k}{n}\right )}^p \log k-\frac{1}{p+1}\log n\rightarrow -\frac{1}{(p+1)^2}$$as $n\rightarrow \infty$

Now we know $$\sum_{k=1}^n \frac{1}{n}{\left ( \frac{k}{n}\right )}^p \log \left (\frac{k}{n} \right )\rightarrow \int_0^1x^p\log x \ dx=-\frac{1}{(p+1)^2}$$

To balance we have to evaluate the limit of $$\left (\sum_{k=1}^n \frac{1}{n}{\left ( \frac{k}{n}\right )}^p -\frac{1}{p+1}\right ) \log n$$ Now observe the sum in brackets is the error of the Riemann sum associated to the Riemann Integral $\displaystyle{\int_0^1x^p dx}$

In the interval $\left [\frac{k}{n},\frac{k+1}{n} \right ]$ if we apply MVT to the function $x^p$ we get $$\left |x^p-\left (\frac{k}{n} \right )^p \right |\leq \left |\left (\frac{k+1}{n} \right )^p -\left (\frac{k}{n} \right )^p\right |=\frac{|p z_k^{p-1} |}{n}$$ for some $z_{k}\in \left [\frac{k}{n},\frac{k+1}{n} \right ] $

So we get $$\sup_{x\in \left [\frac{k}{n},\frac{k+1}{n} \right ]}\left |x^p-\left (\frac{k}{n} \right )^p \right |\leq \frac{p}{n}$$ if $p\geq 1$ Then we have $$\left | \sum_{k=1}^n \frac{1}{n}{\left ( \frac{k}{n}\right )}^p -\frac{1}{p+1}\right |=\left |\sum_{k=0}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}}\left(x^p - \left ( \frac{k}{n}\right )^p \right )dx \right |\leq \frac{p}{n}$$ $$\implies \left (\sum_{k=1}^n \frac{1}{n}{\left ( \frac{k}{n}\right )}^p -\frac{1}{p+1}\right ) \log n\rightarrow 0$$ as $n\rightarrow \infty$ and we are done.

I am really having trouble with the $p<1$ case.

Some help will be very appreciated.


Let $p\geq 0$. For $p=0$ we have $$\sum_{k=1}^n \frac{1}{n}{\left ( \frac{k}{n}\right )}^p \log k-\frac{1}{p+1}\log n\rightarrow -\frac{1}{(p+1)^2}$$ $$\Rightarrow \sum_{k=1}^n \frac{1}{n} \log k-\log(n) = \frac{\log(n!)-n\log(n)}{n} =\frac{\log(n!/n^n)}{n} $$Use Stirling's Formula (the error term can be included, but I'll omit it for brevity): $$ \approx \frac{\log(e^{-n}\sqrt{2\pi n})}{n}\to -1 = -(0+1)^{-2} $$Now suppose $0<p<1$: $$\frac{1}{n^{p+1}}\sum_{k=1}^n k^p \log k-\frac{1}{p+1}\log n $$ $$ =\frac{(p+1)}{n^{p+1}(p+1)}\sum_{k=1}^n k^p \log k-\frac{n^{p+1}\log(n)}{n^{p+1}(p+1)} $$ $$ =\frac{(p+1)\sum\limits_{k=1}^n k^p \log k-n^{p+1}\log(n)}{n^{p+1}(p+1)} $$Note the denominator is strictly increasing and divergent. Then by Stolz-Cesaro, we have $$ =\lim_{n\to\infty}\frac{(p+1)\sum\limits_{k=1}^n k^p \log k-n^{p+1}\log(n)}{n^{p+1}(p+1)} $$ $$ \stackrel{SC}{=}\lim_{n\to\infty}\frac{(p+1)(n+1)^{p}\log(n+1)-\left((n+1)^{p+1}\log(n+1)-n^{p+1}\log(n)\right)}{((n+1)^{p+1}-n^{p+1})(p+1)} $$By the Binomial Theorem, the denominator is $(p+1)^2 n^p +O(n^{p-1})$: $$ \stackrel{BT}{=}\lim_{n\to\infty}\frac{(p+1)(n+1)^{p}\log(n+1)+n^{p+1}\log(n)-(n+1)^{p+1}\log(n+1)}{(n^p +O(n^{p-1}))(p+1)^2} $$ $$ {=}\frac{1}{(p+1)^2}\lim_{n\to\infty}\frac{p(1+n^{-1})^{p}\log(n+1)+n\log(n)-(1+n^{-1})^{p}n\log(n+1)}{1 +O(n^{-1})} $$ $$ {=}\frac{1}{(p+1)^2}\lim_{n\to\infty}\log\left(\frac{(n+1)^{(p+1)(1+n^{-1})^{p}}\cdot n^n}{(n+1)^{n(1+n^{-1})^{p}}}\right) $$ $$ {=}\frac{1}{(p+1)^2}\lim_{n\to\infty}\log\left(\frac{n^n}{(n+1)^{(n-p-1)(1+n^{-1})^{p}}}\right) $$ $$ {=}\frac{1}{(p+1)^2}\log\left(\lim_{n\to\infty}\frac{n^n}{(n+1)^{(n-p-1)(1+n^{-1})^{p}}}\right) $$If you stare at this limit, you will convince yourself it approaches $e^{-1}$ (it can be shown using LHR but this post is long and hideous enough). Then we have $$ {=}\frac{1}{(p+1)^2}\log\left(\lim_{n\to\infty}\frac{n^n}{(n+1)^{(n-p-1)(1+n^{-1})^{p}}}\right) {=}\frac{1}{(p+1)^2}\log\left(e^{-1}\right) = \frac{-1}{(p+1)^2} $$There's probably a nicer way to do this, but that's what I've got!

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    $\begingroup$ This one is getting complicated. Simpler alternatives are available in the comments to question. +1 however for the effort (typing it also). $\endgroup$ – Paramanand Singh May 27 at 17:10
  • $\begingroup$ Oof, next time I'll check before wasting my time typing it up...lesson learned. $\endgroup$ – Integrand May 27 at 17:20
  • $\begingroup$ By the way the question is not trivial and it is not expected that you can come up with a smart solution in limited time. $\endgroup$ – Paramanand Singh May 27 at 17:24
  • $\begingroup$ Thanks a lot for your help. $\endgroup$ – Ignorant Mathematician May 28 at 3:42

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