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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 $x$? –  robjohn Mar 7 '12 at 1:18
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If $\tau$ is number of divisors, it follows from terrytao.wordpress.com/2008/09/23/the-divisor-bound –  sdcvvc Mar 7 '12 at 1:34
    
@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! –  JavaMan Mar 7 '12 at 2:01
    
@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) –  VVV Mar 7 '12 at 12:35
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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 $\frac{\log \tau(n)}{\log n} \to 0$. –  sdcvvc Mar 8 '12 at 2:52

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up vote 3 down vote accepted

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. –  VVV Mar 8 '12 at 11:19

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