Whether or not a series converges does not depend on if $\lim_{n\to\infty}a_n=0$, rather it is the converse that is true.
If a series converges, then $\lim_{n\to\infty}a_n=0$.
But $\lim_{n\to\infty}a_n=0$ does not prove that a series converges.
Your series is an example of one that does not converge and approaches $0$.
Here is my explanation:
If a sum gets closer and closer to some limit, then its partial sums must increase by less and less. Upon reaching the limit, the partial sums no longer increase and we must have the following:$$\lim_{n\to\infty}\sum_{i=1}^{n}{f(i)}=\lim_{n\to\infty}\sum_{i=1}^{n+1}{f(i)}$$Due to the nature of $\infty$.
Furthermore, we have such:
$$\lim_{n\to\infty}\sum_{i=1}^{n}{f(i)}=\lim_{n\to\infty}\sum_{i=1}^{n+1}{f(i)}$$$$\lim_{n\to\infty}f(1)+f(2)+f(3)+\cdots f(n)=\lim_{n\to\infty}f(1)+f(2)+f(3)+\cdots f(n)+f(n+1)$$$$0=\lim_{n\to\infty}f(n+1)$$
One can prove a convergent series has this property, but not that a series with this property is convergent.
This means that a series without this property must be divergent, because I just proved all convergent series have this property, meaning if it doesn't, it can't be convergent.
Also, (fun fact) a series' sum is not dependent on whether it converges.
For example:$$\sum_{n=1}^{\infty}n=-\frac12$$If you don't believe, you can look it up. It even has its own wiki page.
Best Answer
$$ n+n^2\log\frac{n}{n+1} = n-n^2\log\frac{n+1}{n} = n-n^2\log\left(1+\frac{1}{n}\right) $$ So do some asymptotics. As $n \to +\infty$, we have \begin{align} \log\left(1+\frac{1}{n}\right) &= \frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right) \\ n^2\log\left(1+\frac{1}{n}\right) &= n-\frac{1}{2}+o(1) \\ n-n^2\log\left(1+\frac{1}{n}\right) &= \frac{1}{2}+o(1) \end{align}