Answer
$\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ is convergent.
Work Step by Step
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{\ln n}{n^2}\]
\[\Rightarrow a_n=\frac{\ln n}{n^2}\]
$a_n$ is positive term and monotonically decreasing.
So Integral test is applicable.
Let \[I=\int_{1}^{\infty}\frac{\ln x}{x^2}dx\]
\[I=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{\ln x}{x^2}dx\;\;\;\;\;\;\;\;\;\;\ldots(1)\]
Let \[I_1=\int\frac{\ln x}{x^2}dx\]
Substitute $y=\ln x\;\Rightarrow x=e^y$
\[\Rightarrow dy=\frac{1}{x}dx\]
\[\Rightarrow I_1=\int\frac{y}{e^y}dy=\int ye^{-y}dy\]
Using integration by parts
\[\Rightarrow I_1=-ye^{-y}+\int e^{-y}dy\]
\[\Rightarrow I_1=-ye^{-y}-e^{-y}=-(y+1)e^{-y}\]
\[\Rightarrow I_1=-(\ln x+1)e^{-\ln x}=-\frac{\ln x+1}{x}\;\;\;\;\;\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow\infty}\left[-\frac{\ln x+1}{x}\right]_{1}^{t}\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\ln t+1}{t}+\frac{\ln 1+1}{1}\right]\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\ln t+1}{t}+1\right]\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\ln t+1}{t}\right]+1\]
Which is $\displaystyle\frac{\infty}{\infty}$ case, apply L' Hopitals rule
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[-\frac{\frac{1}{t}}{1}\right]+1\]
\[\Rightarrow I=0+1=1\]
Which is finite so $I$ is convergent.
By integral test, $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ is convergent.