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Chapter 9: Problem 19

Determine whether the following series converge. $$\sum_{k=2}^{\infty}(-1)^{k} \frac{k^{2}-1}{k^{2}+3}$$

Short Answer

Expert verified
Answer: The series diverges.

Step by step solution

01

Identify the series type

The series has a factor of \((-1)^k\), which makes it an alternating series. So, we will be applying the Alternating Series Test to check its convergence.
02

Determine if the absolute value of the terms is decreasing

We need to determine if the absolute value of the terms is decreasing: $$|(-1)^k \frac{k^2-1}{k^2+3}| = \frac{k^2-1}{k^2+3}$$ Compare it with its next term: $$ \frac{k^2-1}{k^2+3} < \frac{(k+1)^2-1}{(k+1)^2+3}$$ Now let's perform some algebraic manipulations on this inequality to see if it's true: $$k^2(k+1)^2 + 3(k+1)^2 < k^2k^2 + 3k^2$$ $$k^4 + 2k^3 + k^2 + 3k^2 + 6k + 3 < k^4 + 3k^2$$ Notice that \(k^4\) and \(3k^2\) are there on both sides, so let's remove them: $$2k^3+ k^2 + 6k + 3 < 0$$ Now, it's clear that the inequality is not true for all k, so the absolute value of the terms is not always decreasing.
03

Determine if the limit of the absolute value of terms approaches zero

Now, let's find the limit of the absolute value of the terms as k approaches infinity: $$\lim_{k\to\infty} \frac{k^2-1}{k^2+3}$$ We know that for large values of k, the dominant terms are \(k^2\) in both numerator and denominator: $$\lim_{k\to\infty}\frac{k^2}{k^2} = 1$$ So, the limit of the absolute value of the terms does not approach zero as k approaches infinity.
04

Conclusion

Since neither of the Alternating Series Test conditions were met (the absolute value of the terms is not decreasing, and the limit of the absolute value of terms does not approach zero), we conclude that the given series does not converge.

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