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Let $d(m)$ denote the number of divisors of $m$ and let $N$ be a large integer. Then we have $$\sum_{n \leq N}\frac{d(n)}{n} \geq \left(\sum_{n \leq \sqrt{N}}\frac{1}{n}\right)^{2} \sim \log^{2}N.$$ What prevents me from doing $$\sum_{n \leq N}\frac{d(n)}{n} \geq \left(\sum_{n \leq N^{1/k}}\frac{1}{n}\right)^{k} \sim \log^{k}N$$ for every integer $k$?

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2 Answers 2

Let $N=6000$ and $k=4$.

Then $$\sum_{n\le N} \frac{d(n)}{n} = 48.3654245...$$ while $$ \left( \sum_{n \le N^{1/4}} \frac{1}{n} \right)^4 = 54.564037875...$$

So your second inequality is not true in general.

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The first inequality is: $d(n)$ (th number of divisors of $n$) is at most the number of factorizations $n=ab$ with $a,b\leq\sqrt{N}$, which is clearly true. However, from the RHS of the (suggested) second inequality you'd get rather the number of factorizations $n=a_1a_2\dots a_k$ with $a_k\leq \sqrt[k]{N}$, and this is in general larger than $d(n)$.

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Dear user, Do you mean "... with $a_k \leq N^{1/k}$ ..."? Regards, –  Matt E Nov 29 '12 at 4:25
    
@MattE: thanks, corrected –  user8268 Nov 29 '12 at 9:39

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