<s>[INST] Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ Recently, I encountered a problem about infinite series.
So my question is how to know whether the infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is convergent?
 [/INST] $\sum_{n≥2} \frac{1}{n(\log n)}=\sum_{n≥2} \int_{1}^{\infty} \frac{1}{n^x} dx$
$=\int_{1}^{\infty} \zeta_2(x) dx$   where  [$\zeta_2(x)=\zeta(x)-1$]
So, this is the area under the curve $\zeta(x), x≥1, y≥0$.
Now, for $x>1$,
$\zeta_2(x) > {\frac{1}{2^x} + 2\frac{1}{4^x}+ 4\frac{1}{8^x} +......}$
$=\frac{1}{2}(\frac{1}{2^{x-1}}+\frac{1}{4^{x-1}}+....)$
$=\frac{1}{2^x}(\frac{1}{1-\frac{1}{2^{x-1}}})$
$=\frac{1}{2}(\frac{1}{2^{x-1}-1})$.
So, $\int_{1}^{\infty} \zeta_{2}(x) dx >\int_{1}^{\infty} \frac{1}{2}(\frac{1}{2^{x-1}-1}) dx=\int_{0}^{\infty} \frac{1}{2}(\frac{1}{2^z-1})dz $ which diverges.
This means $\int_{1}^{\infty} \zeta_{2}(x) dx$ also diverges.
And this implies $\sum_{n≥2} \frac{1}{n(\log n)}$ diverges.
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