Use the limit comparison test to determine if the series converges or diverges.
$$\sum _{n=1}^{\mathrm{\infty }}\frac{7\sqrt{n}}{8n^{3/2}+3n-7}$$

Converges
Diverges

Step 1 ：Identify a simpler series to compare with. In this case, we can use $b_n=\frac{1}{n^{1/2}}$, because the highest power of n in the denominator of $a_n$ is $n^{3/2}$.

Step 2 ：Calculate the limit as n approaches infinity of $\frac{a_n}{b_n}$: $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{7\sqrt{n}}{8n^{3/2}+3n-7}}{\frac{1}{n^{1/2}}}$.

Step 3 ：Simplify the above expression: $=\underset{n\to \mathrm{\infty }}{lim}\frac{7n}{8n^{3/2}+3n^{3/2}-7n^{1/2}}=\underset{n\to \mathrm{\infty }}{lim}\frac{7}{8n+3\sqrt{n}-7}$.

Step 4 ：As n approaches infinity, the term $3\sqrt{n}$ and $-7$ become insignificant compared to $8n$. So, the limit simplifies to: $=\underset{n\to \mathrm{\infty }}{lim}\frac{7}{8n}=0$.

Step 5 ：Since the limit is a finite number (not equal to zero), we can conclude that the series $a_n$ and $b_n$ either both converge or both diverge.

Step 6 ：Since the series $b_n=\frac{1}{n^{1/2}}$ is a p-series with p = 1/2 < 1, it diverges.

Step 7 ：Therefore, by the limit comparison test, the original series also diverges. $\boxed{{\displaystyle \text{The series diverges.}}}$