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Calculus 2 : Taylor and Maclaurin Series

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Example Question #71 : Ap Calculus Bc

Which of the following series does not converge?

Possible Answers:

$\sum_{i=0}^{\infty} \frac{(-1)^n}{n}$

$\sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$

$\sum_{i=0}^{\infty} 0.9999999999999^n$

$\sum_{i=0}^{\infty} \frac{n - 1}{ n^3 + 1}$

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

Correct answer:

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

Explanation:

We can show that the series $\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$ diverges using the ratio test.

$L = \lim_{n ->\infty } \frac{a_{n+1}}{a_{n}} = \lim_{n ->\infty } \left[\left(\frac{(n+1)!}{(n+1)^2 cos (n+1)} \div \frac{n!}{n^2 cos (n) }\right)\right]$

$= \lim_{n ->\infty} \frac{n^2(n+1)cos(n)}{(n+1)^2cos(n+1)} = \lim_{n ->\infty} \frac{n^2cos(n)}{(n+1)cos(n+1)}$

n^2 will dominate over (n+1) since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than n^2 , and thus having in the numerator will make the series diverge by the $n^{th}$ limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

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Calculus 2 : Taylor and Maclaurin Series

Study concepts, example questions & explanations for Calculus 2

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Example Question #71 : Ap Calculus Bc

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