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Maple tells me that $\lim_{n \to \infty} n^2 \log^n(1 - \frac{c \log n}{n}) = 0$ for any constant $c$, but I can't find a way to prove it. Any suggestions?

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What have you tried so far? Do you want a proof using the definition of limit, or just evaluating the limit with asymptotics (for instance) is enough of a dimonstration to you? You should make the question a bit more insightful, I guess. –  Andy Oct 20 '10 at 18:37
    
are you taking the logarithm of the $n$th power, or the $n$th power of the logarithm? –  Arturo Magidin Oct 20 '10 at 18:57
    
@Arturo: hakos must mean the latter because the former does not give a limit of zero. –  Qiaochu Yuan Oct 20 '10 at 19:01
    
@Qiaochu: Well, I assume so, but better to check (and to point out that the question is ambiguous as stated...) –  Arturo Magidin Oct 20 '10 at 19:08
    
Thanks guys, I clarified the question. –  hakos Oct 20 '10 at 19:17
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3 Answers 3

up vote 6 down vote accepted

Well $$\log\left(1-c\frac{\log n}{n}\right)=-c\frac{\log n}{n} +O\left(\frac{(\log n)^2}{n^2}\right)$$ and so $$n^2\log\left(1-c\frac{\log n}{n}\right)^n=(-c)^n\frac{(\log n)^n}{n^{n-2}} \left(1+O\left(\frac{\log n}{n}\right)\right)^n.$$ Now $$n\log\left(1+O\left(\frac{\log n}{n}\right)\right)=O(\log n)$$ so that $$\left(1+O\left(\frac{\log n}{n}\right)\right)^n$$ grows as at most a polynomial in $n$. The $n^n$ in the denominator will swamp everything else...

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The Taylor series expansion $- \log (1-x) = \sum_{k \ge 1} \frac{x^k}{k}$ shows that $\log(1 - x) = O(x)$ for small $x$. So $\log \left( 1 - \frac{c \log n}{n} \right) = O \left( \frac{\log n}{n} \right)$, hence the absolute value of the expression in question is bounded by a constant times $\frac{(\log n)^n}{n^{n-2}}$ which rather clearly goes to zero.

This is a good example of why it's more flexible to use big-O notation than to deal directly with the definition of a limit.

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"This is a good example of why it's more flexible to use big-O notation than to deal directly with the definition of a limit." That's basically why I asked if he wanted a proof using the definition! =) –  Andy Oct 20 '10 at 19:38
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This is also a good example of why it is dangerous to use the big-O notation without the utmost care for the constants hidden therein. Here, the absolute value of the expression in question is not a priori bounded by a constant times $\frac{(\log n)^n}{n^{n-2}}$ but only by $c^n\frac{(\log n)^n}{n^{n-2}}$ for some constant $c$. (Fortunately all this is washed away by the denominator $n^{n-2}$.) –  Did Nov 10 '11 at 7:57
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The case $c=0$ is clear, and then for the rest of its values it's enough to consider and calculate $\lim_{n\rightarrow\infty} n^2 \log^n(1 + \frac{|c| \log n}{n})$. Then we have that: $$\lim_{n\rightarrow\infty} n^2 \log^n(1 + \frac{|c| \log n}{n})=\lim_{n\rightarrow\infty} n^2 \left(\frac{|c|\log n}{n}\right)^{n} \left(\left(\frac{\log(1 + \frac{|c| \log n}{n})}{(\frac{|c| \log n}{n})}\right)^{\frac{n}{|c|\log n}}\right)^{|c|\log n}= \lim_{n\rightarrow\infty} n^{2} \left(\frac{|c|\log n}{n}\right)^{n} \frac{1}{e^{\frac{|c|\log n}{2}}}=0.$$

The auxiliary limit i resorted to is $\lim_{x\to0} \left({\frac{\ln(1+x)} {x}}\right)^\frac{1}{x}= \frac{1}{\sqrt{e}}.$

The proof is complete.

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