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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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    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