The problem asks us to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$.

1. Problem Description

The problem asks us to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$.

2. Solution Steps

We can use the limit comparison test to determine the convergence or divergence of the series.

Let $a_n = \frac{\sqrt{2n+1}}{n^2}$. We need to find a series $b_n$ to compare with.

As $n$ becomes large, $2n+1 \approx 2n$, so $\sqrt{2n+1} \approx \sqrt{2n} = \sqrt{2} \sqrt{n}$.

Thus, $a_n \approx \frac{\sqrt{2} \sqrt{n}}{n^2} = \frac{\sqrt{2}}{n^{3/2}}$.

Let $b_n = \frac{1}{n^{3/2}}$.

Now, consider the limit

$$L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{\sqrt{2n+1}}{n^2}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{\sqrt{2n+1}}{n^2} \cdot \frac{n^{3/2}}{1} = \lim_{n \to \infty} \frac{\sqrt{2n+1}}{\sqrt{n^4}} \cdot n^{3/2} = \lim_{n \to \infty} \frac{\sqrt{2n+1}}{\sqrt{n}} = \lim_{n \to \infty} \sqrt{\frac{2n+1}{n}} = \lim_{n \to \infty} \sqrt{2 + \frac{1}{n}} = \sqrt{2+0} = \sqrt{2}.$$

Since $0 < L = \sqrt{2} < \infty$, the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ either both converge or both diverge.

We know that $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ is a p-series with $p = \frac{3}{2} > 1$.

Therefore, the p-series $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges.

Since $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges and the limit $L$ is a positive finite number, the series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$ also converges.

3. Final Answer

Converges