$x_n = \left[1^{1^p}2^{2^p}\cdots (n+1)^{(n+1)^p}\right]-\left[1^{1^p}2^{2^p}\cdots n^{n^p}\right]$ converge?

Question

This problem comes from the Titu Andreescu's book Problems in Real Analysis - Chapter 1, page 9.

Let $p$ be a nonnegative real number. Study the convergence of the sequence

$$x_n = \left[1^{1^p}2^{2^p}\cdots (n+1)^{(n+1)^p}\right]^{1/(n+1)^{p+1}}-\left[1^{1^p}2^{2^p}\cdots n^{n^p}\right]^{1/n^{p+1}}$$

Where $n$ is a positive integer.

Maybe it is useful to know:

$$ \lim_{n \to \infty} \frac{\left(1^{1^p}2^{2^p}\cdots n^{n^p}\right)^{1/n^{p+1}}}{n^{1/(p+1)}} = e^{-1/(p+1)^2}\label{1}\tag{1}$$

Attempt

Just before this exercise, the book solves the case $p =0$ in an example. Using the same reasoning of the example, I did the following:

Define $a_n= \left[1^{1^p}2^{2^p}\cdots n^{n^p}\right]^{1/n^{p+1}}$ and $b_n = a_{n+1}/a_n$, then

$$x_n = a_{n+1}-a_{n} = a_n(b_n-1) = \frac{a_n}{n^{1/(p+1)}}\frac{b_n-1}{\ln b_n}\ln b_n^{n^{1/(p+1)}}$$

But by \ref{1}, we have

$$ \lim_{n \to \infty} \frac{a_n}{n^{1/(p+1)}} = e^{-1/(p+1)^2}$$

And we also have

$$ \lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a_{n+1}}{(n+1)^{1/(p+1)}}\frac{n^{1/(p+1)}}{a_n}\frac{(n+1)^{1/(p+1)}}{n^{1/(p+1)}} = \lim_{n \to \infty} \frac{a_{n+1}}{(n+1)^{1/(p+1)}}\frac{n^{1/(p+1)}}{a_n}\left(1+\frac{1}{n}\right)^{1/(p+1)} = 1$$

Thus,

$$ \lim_{n \to \infty} \frac{b_n-1}{\ln b_n} = \lim_{n \to \infty} \left(\frac{\ln b_n}{b_n-1}\right)^{-1} = \left[\left(\frac{d}{dx}\ln x \right)|_{x=1}\right]^{-1}= 1 $$

Since $b_n \to 1$. So, we just need to analyze the convergence of $\ln b_n^{n^{1/(p+1)}}$, what I couldn't do! However, in the case $p=0$, the book does the following

$$\lim_{n \to \infty} b_n^n = \lim_{n \to \infty} \left(\frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}\right)^n = \lim_{n \to \infty} \frac{(n+1)!}{n!}\frac{1}{(n+1)!^{1/(n+1)}} = \lim_{n \to \infty} \frac{n+1}{(n+1)!^{1/(n+1)}} = e $$

Where we used \ref{1} in the last equality.

Answer 1

We have that

$$x_n = \left[1^{1^p}2^{2^p}\cdots (n+1)^{(n+1)^p}\right]^{1/(n+1)^{p+1}}-\left[1^{1^p}2^{2^p}\cdots n^{n^p}\right]^{1/n^{p+1}}=a_{n+1}-a_n$$

with

$$a_n=\left[1^{1^p}2^{2^p}\cdots n^{n^p}\right]^{1/n^{p+1}}=e^{\frac{\sum_{k=1}^{n}k^p \log k}{n^{p+1}}}\sim cn^{\frac1{p+1}}$$

indeed

$$\frac{\sum_{k=1}^{n}k^p \log k}{n^{p+1}} =\frac1n\sum_{k=1}^{n}\left(\frac kn\right)^p \left(\log \left(\frac kn\right) +\log n\right)=$$

$$=\int_0^1 x^p \log x dx+\frac{\log n}{n^{p+1}}\sum_{k=1}^{n} k^p=\frac{\log n}{p+1}+I+O\left(\frac{\log n}n\right)$$

therefore

$$a_{n+1}-a_n\sim c\cdot \left[(n+1)^{\frac1{p+1}}-n^{\frac1{p+1}}\right]$$

which converges for any $p>0$ indeed let $a=\frac1{p+1}\in (0,1)$

$$(n+1)^a-n^a=n^a(1+1/n)^a-n^a\sim\frac{a}{n^{1-a}} \to 0$$