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.
 A: 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!
