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

Diverges since $b_{n}$ converges but $L = - 1 < 0$

Converges since $b_{n}$ diverges but $L=-1<0$

Converges since $b_{n}$ converges and $L=1>0$

Diverges since $b_{n}$ diverges and $L equals 1 is greater than 0$

검증된 단계별 안내

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.

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.

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 $.

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.

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.

5:35m

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