Convergence or divergence of $\sum_{n=1}^{\infty} (n\cdot \log(\frac{3n+2}{3n-2}) -1)$
For large $n$ $\Rightarrow$ $0< \frac{4}{3n-2}<1$
$$n\cdot \log(\frac{3n+2}{3n-2}) -1 = n\cdot \log(1+\frac{4}{3n-2})-1 = n\cdot(\frac{4}{3n-2} - \frac{16}{2\cdot(3n-2)^2} \ldots) - 1 = \frac{4n}{3n-2} -\frac{8n}{\cdot(3n-2)^2} \ldots - 1 \to \frac43 -1 =\frac13 \neq 0$$ as $n\to \infty$
Hence divergent by Nth term divergence test.
Is this correct?
I don't know what the $N$th term test is, but since$$\lim_{n\to\infty}n\log\left(\frac{3n+2}{3n-2}\right)-1\neq0\tag1,$$the series diverges.
However, your way of proving that we have $(1)$ has a problem. You assume that you can exchande a limit with an infinite sum. It works in this case, but it is not true in general.
You can prove that $\lim_{n\to\infty}n\log\left(\frac{3n+2}{3n-2}\right)=\frac43$ defining $f(x)=\log\left(\frac{3+2x}{3-2x}\right)$ and noticing that\begin{align}\lim_{n\to\infty}n\log\left(\frac{3n+2}{3n-2}\right)&=\lim_{n\to\infty}\frac{f\left(\frac1n\right)}{\frac1n}\\&=f'(0)\\&=\frac43.\end{align}