Is $\lim_{n\to +\infty} \frac{\sqrt{n+1}}{2n^2+1}$ convergent? I don't know how to simplify this expression to apply the ratio test to determine if the sequence converges or diverges
$$\lim_{n\to +\infty} \frac{\sqrt{n+1}}{2n^2+1}$$

Estimate sequence terms above and below 
$$\frac{1}{2(n+1)^{\frac{3}{2}}}=\frac{\sqrt{n+1}}{2(n+1)^2} \leqslant  \frac{\sqrt{n+1}}{2n^2+1} \leqslant \frac{2\sqrt{n}}{2n^2}=\frac{\sqrt{n}}{n^2}=\frac{1}{n^{\frac{3}{2}}}.$$
Hence $$\frac{\sqrt{n+1}}{2(n+1)^2}=O({n^{-\frac{3}{2}}}), \;\;\;n\to\infty.$$
