We are asked to determine whether the series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$ converges or diverges.

1. Problem Description

We are asked to determine whether the series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$ converges or diverges.

2. Solution Steps

We can use the limit comparison test to determine the convergence of the series. Let $a_n = \frac{\sqrt{2n+1}}{n^2}$. We want to compare $a_n$ with a simpler series $b_n$. As $n$ gets large, $2n+1 \approx 2n$, so $\sqrt{2n+1} \approx \sqrt{2n} = \sqrt{2} \sqrt{n}$. Therefore, $a_n \approx \frac{\sqrt{2n}}{n^2} = \frac{\sqrt{2} \sqrt{n}}{n^2} = \frac{\sqrt{2}}{n^{3/2}}$. So we can choose $b_n = \frac{1}{n^{3/2}}$.

Now, let's compute the limit:

$$$$

\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 n^{3/2} = \lim_{n \to \infty} \frac{\sqrt{2n+1}}{\sqrt{n^4}} \cdot \sqrt{n^3} = \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}

$$$$

Since the limit is $\sqrt{2}$, which is a finite, positive number, the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ either both converge or both diverge. We know that the series $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ is a $p$-series with $p = \frac{3}{2} > 1$, so it converges. Therefore, the series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$ also converges.

3. Final Answer

The series converges.