It is easy to derive that $h_{n}=\frac{2(n+1)^{n+1}}{n^{n}(2 n+1)}$. Consider the function $g(x)$ defined by:
$$g(x)=\ln 2+(x+1) \ln (x+1)-x \ln x-\ln (2 x+1)$$

Then, for $0<x<+\infty$, we have
$$\begin{array}{c} g^{\prime}(x)=\ln (x+1)-\ln x-\frac{2}{2 x+1} \\ g^{\prime \prime}(x)=\frac{1}{x+1}-\frac{1}{x}+\frac{4}{(2 x+1)^{2}}=-\frac{1}{x(x+1)(2 x+1)^{2}}<0 \end{array}$$

Therefore, $g^{\prime}(x)$ is monotonically decreasing on $(0,+\infty)$.
Since $\lim _{x \rightarrow \infty} g^{\prime}(x)=\lim _{x \rightarrow \infty} \ln \frac{x+1}{x}-\lim _{x \rightarrow \infty} \frac{2}{2 x+1}=0$, and knowing that $g^{\prime}(x)$ is positive on $(0,+\infty)$, it follows that $g(x)$ is monotonically increasing on $(0,+\infty)$. Therefore, when $n$ is a positive integer, $h_{n}=\mathrm{e}^{g(n)}$ is a strictly increasing sequence, i.e., $h_{1}<h_{2}<\cdots<h_{n}<h_{n+1}$.