We naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is \(1/n^2\text{.}\) Knowing that \(\ds \infser \frac1{n^2}\) converges, we attempt to apply the Limit Comparison Test:

\begin{align*}
\lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^2-n+1)}{1/n^2} \amp = \lim_{n\to\infty}\frac{n^2(\sqrt n+3)}{n^2-n+1}\\
\amp = \infty \text{ (Apply L'Hospital's Rule) } \text{.}
\end{align*}

Theorem 9.3.9 part (3) only applies when

\(\ds\infser b_n\) diverges; in our case, it converges. Ultimately, our test has not revealed anything about the convergence of our series.

The problem is that we chose a poor series with which to compare. Since the numerator and denominator of the terms of the series are both algebraic functions, we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator.

The dominant term of the numerator is \(n^{1/2}\) and the dominant term of the denominator is \(n^2\text{.}\) Thus we should compare the terms of the given series to \(n^{1/2}/n^2 = 1/n^{3/2}\text{:}\)

\begin{align*}
\lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^2-n+1)}{1/n^{3/2}} \amp = \lim_{n\to \infty} \frac{n^{3/2}(\sqrt n+3)}{n^2-n+1}\\
\amp = 1 \text{ (Apply L'Hospital's Rule) } \text{.}
\end{align*}

Since the \(p\)-series \(\ds\infser \frac1{n^{3/2}}\) converges, we conclude that \(\ds\infser \frac{\sqrt{n}+3}{n^2-n+1}\) converges as well.