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Multiple Choice
Which of the following series can be shown to converge by using the ratio test?
A
B
C
D
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1
Step 1: Recall the ratio test. The ratio test states that for a series \( \sum_{n=1}^{\infty} a_n \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \), then: (a) if \( L < 1 \), the series converges absolutely, (b) if \( L > 1 \), the series diverges, and (c) if \( L = 1 \), the test is inconclusive.
Step 2: Apply the ratio test to \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). Here, \( a_n = \frac{1}{n^2} \). Compute \( \frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}} = \frac{n^2}{(n+1)^2} \). Take the limit as \( n \to \infty \): \( \lim_{n \to \infty} \frac{n^2}{(n+1)^2} \). This simplifies to \( 1 \), so the ratio test is inconclusive for this series.
Step 3: Apply the ratio test to \( \sum_{n=1}^{\infty} \frac{1}{n} \). Here, \( a_n = \frac{1}{n} \). Compute \( \frac{a_{n+1}}{a_n} = \frac{\frac{1}{n+1}}{\frac{1}{n}} = \frac{n}{n+1} \). Take the limit as \( n \to \infty \): \( \lim_{n \to \infty} \frac{n}{n+1} \). This simplifies to \( 1 \), so the ratio test is inconclusive for this series as well.
Step 4: Apply the ratio test to \( \sum_{n=1}^{\infty} \frac{1}{n!} \). Here, \( a_n = \frac{1}{n!} \). Compute \( \frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} = \frac{n!}{(n+1)!} = \frac{1}{n+1} \). Take the limit as \( n \to \infty \): \( \lim_{n \to \infty} \frac{1}{n+1} \). This simplifies to \( 0 \), which is less than \( 1 \). Therefore, the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges by the ratio test.
Step 5: Apply the ratio test to \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \). Here, \( a_n = \frac{(-1)^n}{n} \). Compute \( \frac{a_{n+1}}{a_n} = \frac{\frac{(-1)^{n+1}}{n+1}}{\frac{(-1)^n}{n}} = \frac{n}{n+1} \). Take the limit as \( n \to \infty \): \( \lim_{n \to \infty} \frac{n}{n+1} \). This simplifies to \( 1 \), so the ratio test is inconclusive for this series.