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Is there a way to show that $d(n)$, which counts the number of divisors of $n$ is non-increasing?

I'm trying to use the Cauchy condensation test to show that $\sum_{n\ge{2}}\frac{d(n)}{n\log^2n}$ is divergent. In order to do this, I have to show that $\frac{d(n)}{n\log^2n}$ is non increasing,

If I can show that $d(n)$ is non-increasing then I am done, but I don't know where to start with this. Any help is appreciated.

Martin Sleziak
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Pablo
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    What do you mean by "non-increasing"? Just the negation of increasing? Or $n<m \Rightarrow d(n)\geqslant d(m)$ ? These two things are different: the former is true whereas the latter is false. – Taladris Sep 16 '14 at 06:14
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    Just the negation of increasing. – Pablo Sep 16 '14 at 06:18
  • Showing that $\frac{d(n)}{n\log^2n}$ is non-increasing is vastly different from showing that $d(n)$ is non-increasing. The latter isn't so, but that has no bearing on the former. – Steven Stadnicki Sep 16 '14 at 06:24
  • would you mind flagging this for deletion? – Pablo Sep 16 '14 at 06:25
  • I guess you should ask "convergence or divergence" of the series $\sum \frac{d(n)}{n\log^2 n}$ – Sungjin Kim Sep 16 '14 at 06:26

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I assume that by 'non-increasing', you mean 'monotonically decreasing'. (You said in the comment that you wanted to apply Cauchy condensation test and Wikipedia's Cauchy condensation test article uses such term)

Well, the problem is that $\frac{d(n)}{n\log^2 n}$ is not monotonically decreasing since $$\frac{d(p)}{p\log^2p}=\frac{2}{p\log^2p}<\frac{3}{(p+1)\log^2(p+1)}\le\frac{d(p+1)}{(p+1)\log^2(p+1)}$$ for all sufficiently large prime $p$. So you cannot use that test.

The divergence can be shown by estimating the sum assymptotically. Let $$C(x):=\sum_{k\le x}c_n$$ Then \begin{align} \sum_{n=a}^{b}c_nf(n)&=\sum_{n=a}^{b}f(n)(C(n)-C(n-1))\\ &=\sum_{n=a}^{b-1}C(n)(f(n)-f(n+1))+f(b)C(b)-f(a)C(a-1)\\ &=f(b)C(b)-f(a)C(a-1)-\sum_{n=a}^{b-1}C(n)\int_n^{n+1}f'(x)\,\mathrm{d}x\\ &=f(b)C(b)-f(a)C(a-1)-\sum_{n=a}^{b-1}\int_n^{n+1}f'(x)C(x)\,\mathrm{d}x\\ &=f(b)C(b)-f(a)C(a-1)-\int_a^bf'(x)C(x)\,\mathrm{d}x \end{align} Now put $$f(x)=\frac{1}{x\log^2x}, c_n=d(n)$$ Then we have $$\sum_{n=2}^m\frac{d(n)}{n\log^2n}=\frac{1}{m\log^2m}\sum_{n\le m} d(n)-\frac{1}{2\log^22}+\int_2^m\frac{2+\log x}{x^2\log^3x}\sum_{n\le x}d(n)\,\mathrm{d}x$$ There is an estimation on the sum of number divisors using the hyperbola method.(see this) That is,$$\sum_{n\le x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt{x})$$ Now plugging this on the above sum, we get$$\sum_{n=2}^m\frac{d(n)}{n\log^2n}=\log\log m+O(1)$$ which is divergent.

Community
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karvens
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How could $d(n)$ be non-increasing? $3$ has $2$ divisors, $4$ has $3$ divisors, meaning that $$d(3)=2\\ d(4)=3\\ d(5)=2$$ meaning $d$ is neither non-decreasing nor non-increasing.

5xum
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    I understand, but I am trying to use the Cauchy condensation test in showing that $\sum_{n\ge{2}}\frac{d(n)}{n\log^2n}$ is divergent, and to do so I must show that $\frac{d(n)}{n\log^2n}$ is non increasing. Do you have any suggestions? – Pablo Sep 16 '14 at 06:18
  • I mean that the sequence is non-increasing. But I think you are right. It cannot be increasing. – Pablo Sep 16 '14 at 06:23