Check convergence of series I want to check the convergence of the following series $$\sum_{n=0}^{\infty}\left (\frac{3+n^3}{4-n+n^2-3n^3}\right )^n \ \\ \sum_{n=0}^{\infty}\left (\frac{3+n^2}{4+2n^2}\right )^n \\ \sum_{n=1}^{\infty}\left (\sqrt{2n}-\sqrt{n}\right )  $$
I have done the following :
$$\sum_{n=0}^{\infty}\left (\frac{3+n^3}{4-n+n^2-3n^3}\right )^n \rightarrow \lim_{n\rightarrow \infty}\left |\frac{3+n^3}{4-n+n^2-3n^3}\right |=\left |-\frac{1}{3}\right |=\frac{1}{3}<1 \text{ Convergence } \\ \sum_{n=0}^{\infty}\left (\frac{3+n^2}{4+2n^2}\right )^n \rightarrow \lim_{n\rightarrow \infty}\left |\frac{3+n^2}{4+2n^2}\right |=\frac{1}{2}<1 \text{ Convergence }\\ \sum_{n=1}^{\infty}\left (\sqrt{2n}-\sqrt{n}\right ) \rightarrow \sqrt{2n}-\sqrt{n} = \sqrt{2}\sqrt{n}-\sqrt{n} =(\sqrt{2}-1)\cdot \sqrt{n} \geq \frac{1}{4}\sqrt{n} \\ \text{ since the series of } \sqrt{n} \text{ diverges we get that also the original series diverges}$$ Is everything correct ?

The limit of the general term is,  after your work,  $\lim_{n\to\infty}(\sqrt2-1)\sqrt n=\infty$.
Thus since it doesn't go to zero,  the series diverges (sometimes called the divergence test).
