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