Doubt – used inequality, $1/(n^2 \log n) < 1/ n^2$ will be true only for $n > 10$ as $\log n < 1$ for $1 < n < 10 $. but here summation is running from $n= 2$ to infinite
please check whether the solution which is arrived is correct with the correct procedure.
Best Answer
Your idea is right, we have that
$$\frac{\frac1{n^2 \log n}}{\frac1{n^2}}=\frac1{\log n} \to 0$$
and therefore by LCT the given series converges.
Note that the initial values are not relevant here since we are interested to the behaviour for $n$ larg and also the inequality $1/(n^2 \log n) < 1/ n^2$ is not an issue when we use LCT.