91Ó°ÊÓ

First, we note down the provided series as \(\sum_{n=1}^{\infty} \frac{2 n^{2}-1}{3 n^{5}+2 n+1}\)

We then identify a simpler series that we can compare it with. This is typically the series that you get by taking the highest degree terms from the numerator and denominator. This would be \(\frac{n^{2}}{n^{5}} = \frac{1}{n^{3}}\), so our comparison series is \(\sum_{n=1}^{\infty} \frac{1}{n^{3}}\)

The Limit Comparison Test states that if the limit as n approaches infinity of the ratio of the nth term of the given series and the nth term of the easier series is a positive finite number, then both series behave the same way (converge or diverge). Our simpler series \(\sum_{n=1}^{\infty} \frac{1}{n^{3}}\) is a well-known p-series that converges as the exponent on n is greater than 1. The limit is computed as follows: \(\lim_{n\to \infty} (\frac{a_n}{b_n}) = lim_{n\to\infty} \frac{(2 n^{2}-1)/(3 n^{5}+2 n+1)}{1/n^{3}} = \lim_{n\to\infty} \frac{2 n^{2}-1}{(3 n^{5}+2 n+1) * n^{3}} = \lim_{n\to\infty} \frac{2 - 1/n^{2}}{3 n + 2/n^{4} + 1/n^{5}}\). Considering the leading coefficients as n approaches infinity, it equals to \(\frac{2}{3}\), a positive finite number.

Since the limit comparison test resulted in a positive finite number, both series (the given series and the simpler series) behave the same way. The simpler series converges (since it’s a p-series with p > 1). Therefore, the given series also converges.