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

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 \(\frac{a_n}{b_n}\): \(\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{7\sqrt{n}}{8n^{3/2} + 3n - 7}}{\frac{1}{n^{1/2}}}\).

Step 3 ：Simplify the above expression: \(= \lim_{n \to \infty} \frac{7n}{8n^{3/2} + 3n^{3/2} - 7n^{1/2}} = \lim_{n \to \infty} \frac{7}{8n + 3\sqrt{n} - 7}\).

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

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. \(\boxed{\text{The series diverges.}}\)