Join thousands of students who trust us to help them ace their exams!Watch the first video
Multiple Choice
Which of the following series can be shown to converge by using the ratio test?
A
B
C
D
Verified step by step guidance
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.