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A problem from "Problems in Real Analysis - Advanced Calculus on the Real Axis":

Let $p$ be a nonnegative real number. Study the convergence of the sequence $(x_n)_{n\ge1}$ defined by $$x_n=\left(1^{1^p}\cdot2^{2^p}\cdots(n+1)^{(n+1)^p}\right)^{1/(n+1)^{p+1}}-\left(1^{1^p}\cdot2^{2^p}\cdots n^{n^p}\right)^{1/n^{p+1}}.$$

If $p=0$, it is already proved in the book that $x_n$ converges to $\dfrac1e$. So just consider the case that $p>0$. If the sequence converges, I can prove by Stolz-Cesàro Theorem that the sequence must converge to $0$. But I don't know how to determine which $p$ gives convergent sequences.

Please don't give complete solution, just helpful hints. Thanks in advance.

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2 Answers 2

HINT : Let $ a_n = \left( 1^{1^p} \cdot 2^{2^p} \cdots n^{n^p} \right)^{1/n^{p+1}}$ so that $x_n = a_{n+1}- a_n $. Now study the asymptotics of $\log a_n$ using the Euler-Maclaurin formula.

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I didn't know what Euler-Maclaurin formula is. I already googled it, but I don't have a slightest clue of how to apply it here. Can you give more elementary hint? –  dkdsj93 Oct 5 '11 at 2:09
    
@3423847238 I'm sorry, I don't have more a more elementary method. Some of these types of questions are designed to be solved by special tricks and clever repeated applications of some theorems like Caesaro-Stoltz, but a general powerful method is Euler-Maclaurin. You should learn about it. The way to apply it here is simple: $\log a_n$ is a sum whose value we can estimate very well if we take enough terms of the Euler-Maclaurin estimate. You can then use that estimate to estimate $a_{n+1} - a_n$. Sorry about this late reply by the way. –  Ragib Zaman Oct 9 '11 at 21:09

Use $\sum_{k=1}^n k^p \log(k) = \frac{\mathrm{d}}{\mathrm{d} p} H_n^{(-p)}$, as well as series expansion for harmonic numbers $H_n^{(-p)}$ of order $-p$, using $H_n^{(-p)} = \zeta(-p) - \zeta(-p, n+1)$.

I am getting zero value for the limit, as you claimed.

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3  
Somehow I get the feeling that calling in the Hurwitz zeta function here is mosquito-nuking... –  J. M. Oct 3 '11 at 16:14
    
This seems to be completely out of my reach.. –  dkdsj93 Oct 5 '11 at 2:10

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