# Evaluate the limit for $p<1$

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.

• – Paramanand Singh May 27 at 17:04
• The limit for the expression after "to balance..." you can use the answer math.stackexchange.com/a/149174/72031 – Paramanand Singh May 27 at 17:08
• This result seems pretty useful. – Ignorant Mathematician May 28 at 3:46
• Yes the result essentially gives us how fast the integral can be estimated by its Riemann sum. It works for typical functions except periodic ones on an interval of length equal to a period. – Paramanand Singh May 28 at 5:58

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 $$0: $$\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!