Answer
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n^{p}}
$$
The given series converges if $p \gt 1$.
Work Step by Step
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n^{p}}
$$
Apply the Integral Test to the series when $p=1$
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n}
$$
The function
$$
f(x)=\frac{\ln x}{x}
$$
is positive and continuous for $x\geq 2$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{1- \ln x}{x^{2}}
$$
So, $f^{'}(x) \lt 0 $ for $x \gt e$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{2}^{\infty} \frac{\ln x}{x} d x &=\lim\limits_{t \to \infty } \int_{2}^{t } \frac{\ln x}{x} d x \\
& \quad\quad\quad \left[\text { use integration by substituting }\right] \\
&\quad\quad\quad \left[ u=\ln x, \quad\quad d u= \frac{dx}{x} \right] , \text { then } \\
&=\lim\limits_{t \to \infty } \int_{2}^{t } u d u\\
&=\lim\limits_{t \to \infty } [\frac{u^{2} }{2} ]_{1}^{t} \\
&=\infty \\
\end{aligned}
$$
(a) So, the series diverges when $p=1$
(b) If $p \ne 1$
Apply the Integral Test to the series when $p\ne 1$
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n^{p}}
$$
The function
$$
f(x)=\frac{\ln x}{x^{p}}
$$
is positive and continuous for $x\geq 2$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{x^{p-1}(1- \ln x)}{x^{2p}}
$$
So, $f^{'}(x) \lt 0 $ for $x \gt e$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{2}^{\infty} \frac{\ln x}{x^{p}} d x &=\lim\limits_{t \to \infty } \int_{2}^{t } \frac{\ln x}{x^{p}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u= \ln x, \quad\quad dv= x^{-p}dx }\\ {d u= \frac{dx}{x} , \quad\quad v= \frac{x^{1-p}}{1-p} }\end{array}\right] , \text { then }\\
&=\lim\limits_{t \to \infty } [\frac{x^{1-p} \ln x}{1-p} -\frac{1}{1-p } \int_{2}^{t } x^{-p} d x ]\\
&=\lim\limits_{t \to \infty } [\frac{x^{1-p} \ln x}{1-p} -\frac{x^{-p+1}}{(1-p)^{2} }] \\
&=\left[\frac{x^{-p+1}}{(-p+1)^{2}}[-1+(-p+1) \ln x]\right]_{2}^{\infty}\\
\end{aligned}
$$
we obtain that if $p \gt 1$ the integral is convergent , therefore, the series converges if $p \gt 1$.
