Convergence/Divergence of some series

I thought it might be instructive to present an approach that relies on the elementary inequality

$$\log(1+x)\le x \tag1$$

and straightforward arithmetic. To that end, we now proceed.

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From $(1)$ we have

$$\begin{align} \frac{1}{n\log(n)}&\ge \frac{\log\left(1+\frac1n\right)}{\log(n)}\\\\ &\ge \log\left(1+\frac{\log\left(1+\frac1n\right)}{\log(n)}\right)\\\\ &=\log\left(\frac{\log(n+1)}{\log(n)}\right)\tag 2 \end{align}$$

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We can now exploit a telescoping series by using $(2)$ to write

$$\begin{align} \sum_{n=2}^N \frac{1}{n\log(n)}&\ge \sum_{n=2}^N \log\left(\frac{\log(n+1)}{\log(n)}\right)\\\\ &=\sum_{n=2}^N \left(\log(\log(n+1))-\log(\log(n)) \right)\\\\ &=\log(\log(N+1))-\log(\log(2))\tag 3 \end{align}$$

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Finally, letting $N\to \infty$ in $(3)$ yields the coveted result

$$\lim_{N\to \infty}\sum_{n=2}^N \frac{1}{n\log(n)}=\infty$$

and the series of interest diverges.

Your comparison goes in the wrong direction: You need $\dfrac 1 {n\log n} \ge \text{something.}$

Alternatively, you can use an integral test: $$ \int_2^\infty \frac{dx}{x\log x} = \int_2^\infty \left(\frac 1 {\log x} \right) \left( \frac{dx} x\right) = \int_{\log 2}^\infty \frac 1 u\, du = +\infty. $$

Integral test

$$ \int_{2}^{\infty}\frac{1}{x\log(x)}\mathbb{d}x $$

$$ u=\log(x) $$

$$ \mathbb{d}u=\frac{1}{x}\mathbb{d}x $$

$$ \int_{\log(2)}^{\infty}\frac{1}{u}\mathbb{d}u=\lim_{a\to\infty}\left[\log|a|-\log|\log(2)|\right]=\infty $$

So the sum must diverge as well.