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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 $k$?

2 Answers 2

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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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    @MattE: thanks, corrected2012-11-29
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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.