What Will be the p value for this converging series? For what values of $p$, the following series converges:
$$
\sum_{n\ge2}^{} \frac{1}{n(\log n)^{p}}
$$
Let
$$a_{n}=\frac{1}{n(\log n)^{p}} \;\;\;\;\;\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;\;\;\;\;\;\;\;a_{n+1}=\frac{1}{(n+1)[\log (n+1)]^{p}}$$
$$\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{n \times(\log n)^{p}}{(n+1)\{\log (n+1)\}^{p}}\right|=L(\operatorname{say})$$
$given$, $summation$ is $convergence$
$so$,L must be $<1$
How to proceed further
 A: cauchy's consideration Test:
If $f(x)$ is a positive, Monotonically decreasing function of
$n$ then for $\ n \in \mathbb{N}$,
the two infinite series $\sum f(x)$ and
$\sum \operatorname{a^{n}}f\left(a^{n}\right)$ converge
or diverge together $(a>1)$
using this test for $\sum_{n=2}^{\infty} \frac{1}{n(\log x)^{p}}$
$
\operatorname{Let} f(n)=\frac{1}{n \cdot[ \log n]^{p}}
$
$\operatorname{case} 1$,
$p$ >$0$
clearly it is decreasing function of $n$
we can use
c.c Test  Here,
\begin{aligned}
 \quad a^{n} f\left(a^{n}\right) &=a^{n} \times \frac{1}{a^{n} \cdot\left(\log a^{n}\right)^{p}} \\
&=\frac{1}{(n \log a)^{p}}=\frac{1}{n^{p} \times \log a^{p}}
\end{aligned}
\begin{aligned}
=\frac{1}{k} \cdot \frac{1}{n^{p}}
\end{aligned}
From p series test we know,
$\sum_{n=1}^{\infty} \frac{1}{n^{p}}\left\{\begin{array}{ll}\text { convergent } & p>1 \\ \text { divergent } & p \leqslant 1\end{array}\right.$
so for this series if $p>1$ : convergent
$case 2$,
for
$p \leqslant 0$,
$\left(\frac{1}{n(\log n)^{p}} \geq \frac{1}{n})\right.$
this is also divergent
According to question required Condition: $P>1$
