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I tried using this inequality: $\tau (n) < (\frac{2}{\log 2} \log n)^{2^{x}}n^{1/x}$ which gives : $$\frac{\log \tau (n)}{\log n} < \frac{(2^{x})\log(2)\log\log n -x\log n}{(\log \log 2) \log n} $$

however then the limit will approach infinity so it isnt useful.

According to the text, $\lim_{n\to\infty}\frac{\log \tau (n)}{\log n}$ is supposed to be 0. How does one see that? (tau is the number of divisor function, $x \in \mathbb{R}_{>0}$)

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    What is this $\tau$ of which you speak? And then, what is $x$?2012-03-07
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    If $\tau$ is number of divisors, it follows from http://terrytao.wordpress.com/2008/09/23/the-divisor-bound/2012-03-07
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    @sdcvvc: I learned item #4 appearing in Tao's blog post from a colleague, and was about to copy the argument (which I have typed in an expository paper). Thanks for saving me the time!2012-03-07
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    @JavaMan could you point out to me which argument from the blog you were about to copy? It's hard for me to see. Thanks. (Also to sdcvvc and robjohn)2012-03-07
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    Its the argument that $\tau(n) \leq C_{\epsilon} n^{\epsilon}$ for every $\epsilon > 0$. I meant a proof to equation (3), and not 4.2012-03-07
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    @JavaMan From $\tau(n) \le C_{\epsilon}n^{\epsilon}$ (which the blog proves) for every $\epsilon > 0$, I suppose you mean putting: $\log \tau(n) $ $\le \epsilon \log (C_{\epsilon} n)$, but that would go to $\epsilon$ and not 0? Anyhow thank you very much, please do put your next comment as an answer so I can give you credit.2012-03-07
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    ( I mean to say that then the quotient: $\frac{\epsilon log(C_{\epsilon} n) }{log(n)}$ would go to $\epsilon$ )2012-03-07
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    VVV: If you prove that for all $\epsilon > 0$ the limit is at most $\epsilon$, it follows that the limit is 0. Your inequality $\tau(n) < (\frac{2 \log n}{\log 2})^{2^x} n^{1/x}$ proves that as well, assuming it is true for all $x>0$. $$\frac{\log \tau(n)}{\log n} \leq \frac{2^x \log(\frac{2\log n}{\log 2})+\frac{1}{x} \log n}{\log n} =\frac{2^x \log (2 \frac{2\log n}{\log 2})}{\log n}+\frac{1}{x} \to \frac{1}{x}$$ therefore $\frac{\log \tau(n)}{\log n} \to 0$.2012-03-08
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    Ah, now I see where I went wrong!! Thanks sdcvvc.2012-03-08

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Perhaps an easier argument to follow is the one given in Hardy and Wright to prove and discuss Theorem 315 in Chapter 18. Everything on pages 343 ($=7^3$) to 346 in the 6th edition is of some interest here.

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    I will have a look at it, thank you Gerry Myerson.2012-03-08