Answer
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{2}^{\infty} \frac{\ln x}{x^{3}} d x =\frac{2\ln 2+1}{16} .
$$
So, the series converges
Work Step by Step
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}}= \frac{\ln 2}{2^{3}}+ \frac{\ln 3}{3^{3}}+\frac{\ln 4}{4^{3}}+...
$$
Apply the Integral Test to the series
$$
\sum_{n=2}^{\infty} \frac{\ln n}{n^{3}}
$$
The function
$$
f(x)=\frac{\ln x}{x^{3}}
$$
is positive and continuous for $x\gt 2$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{1-3\ln x}{x^{4}}.
$$
So, $f^{'}(x) \lt 0 $ for $x\gt 2$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{2}^{\infty} \frac{\ln x}{x^{3}} d x &=\lim\limits_{t \to \infty } \int_{2}^{t } \frac{\ln x}{x^{3}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=\ln x, \quad\quad dv= \frac{dx}{ x^{3} } }\\ {d u= \frac{dx}{x }, \quad\quad v= - \frac{1}{2x^{2}} }\end{array}\right] , \text { then }\\
&=\lim\limits_{t \to \infty } [-\frac{2\ln x+1}{4x^{2}}]_{2}^{t} \\
&=\lim\limits_{t \to \infty } [-\frac{2\ln t+1}{4t^{2}}]+\frac{2\ln 2+1}{16} \\
& \quad\quad\quad\left[\text { by using L'Hospital's rule as follows } \right] \\
& =\lim\limits_{t \to \infty } [-\frac{2\frac{1}{t}}{8t}]+\frac{2\ln 2+1}{16} \\
&= \lim\limits_{t \to \infty } [-\frac{1}{4t^{2}}]+\frac{2\ln 2+1}{16} \\
&=\frac{2\ln 2+1}{16} .
\end{aligned}
$$
So, the series converges