Ytukyg

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.

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

$$frac{sum_{k=1}^{n}k^p log k}{n^{p+1}} =frac1nsum_{k=1}^{n}left(frac knright)^p left(log left(frac knright) +log nright)=$$

$$=int_0^1 x^p log x dx+frac{log n}{n^{p+1}}sum_{k=1}^{n} k^p=frac{log n}{p+1}+I+Oleft(frac{log n}nright)$$

therefore

$$a_{n+1}-a_nsim ccdot left[(n+1)^{frac1{p+1}}-n^{frac1{p+1}}right]$$

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

$$(n+1)^a-n^a=n^a(1+1/n)^a-n^asimfrac{a}{n^{1-a}} to 0$$