Use the Limit Comparison Test to determine if the following series converges.
${\sum }_{n=1}^{\infty }\frac{n^2+1}{n}$

• A. Diverges since $b_{n}$ converges but
• B. Converges since $b_{n}$ diverges but $L=-1<0$
• C. Converges since $b_{n}$ converges and $L=1>0$
• D. Diverges since $b_n$ diverges and $L=1>0$

Answer: (D)

Verified step by step guidance

1. Step 1: Recall the Limit Comparison Test. This test is used to determine the convergence or divergence of a series by comparing it to another series whose behavior is known. Specifically, if \( a_n \) and \( b_n \) are positive sequences, and \( \lim_{n \to \infty} \frac{a_n}{b_n} = L \), where \( L \) is a finite positive number, then both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
2. Step 2: Identify \( a_n \) and \( b_n \) for the given series. Here, \( a_n = \frac{n^2 + 1}{n} \). To simplify \( a_n \), divide each term in the numerator by \( n \): \( a_n = n + \frac{1}{n} \). Choose \( b_n = n \) as the comparison series because it is simpler and its behavior (divergence) is well-known.
3. Step 3: Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \). Substitute \( a_n \) and \( b_n \) into the ratio: \( \frac{a_n}{b_n} = \frac{n + \frac{1}{n}}{n} \). Simplify the expression: \( \frac{a_n}{b_n} = 1 + \frac{1}{n^2} \). As \( n \to \infty \), \( \frac{1}{n^2} \to 0 \), so \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \).
4. Step 4: Interpret the result of the limit. Since \( L = 1 \) is a finite positive number, the Limit Comparison Test tells us that \( \sum a_n \) and \( \sum b_n \) will have the same behavior. Now, examine \( \sum b_n = \sum n \). This series diverges because the terms \( n \) do not approach zero and grow without bound.
5. Step 5: Conclude the behavior of \( \sum a_n \). Since \( \sum b_n \) diverges and \( L = 1 > 0 \), the series \( \sum a_n = \sum \frac{n^2 + 1}{n} \) also diverges by the Limit Comparison Test.