Monthly Archives: October 2020

Solutions

Exercise 7.2.1

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Exercise statement

Is the series \sum_{n=1}^\infty (-1)^n convergent or divergent? Justify your answer. Can you now resolve the difficulty in Example 1.2.2?

Hints

None.

How to think about the exercise

This is a straightforward exercise.

Model solution 1

Let (a_n)_{n=1}^\infty be the sequence defined by a_n := 1 for all n \geq 1. This sequence is non-negative and decreasing, since a_n = 1 \geq 0 and a_n = 1 \geq 1 = a_{n+1} for every n \geq 1. We also have \lim_{n\to\infty} a_n = 1 \ne 0. Thus we can apply the alternating series test (Proposition 7.2.12) to conclude that the series \sum_{n=1}^\infty (-1)^n a_n = \sum_{n=1}^\infty (-1)^n diverges.

The reasoning in Example 1.2.2 is not valid because the series does not converge, so variable S is not a real number and thus cannot be manipulated in the usual way using the laws of algebra.

(Thanks to William for this solution.)

Model solution 2

The series \sum_{n=1}^\infty (-1)^n diverges by the zero test, Corollary 7.2.6. Indeed, the limit \lim_{n\to\infty} (-1)^n does not exist as the sequence ((-1)^n)_{n=1}^\infty = (-1,1,-1,1,-1,\ldots) oscillates between -1 and 1. To prove that \lim_{n\to\infty} (-1)^n does not exist, let \varepsilon := 1. Then no matter how big N \geq 1 is, we can pick j := N and k:=N+1. We have |(-1)^j - (-1)^k| = 2 > \varepsilon. Thus the sequence is not Cauchy, so by Theorem 6.4.18 it is not convergent. (Alternatively, we could have used Lemma 6.5.2.)

The reasoning in Example 1.2.2 is not valid because the series does not converge, so variable S is not a real number and thus cannot be manipulated in the usual way using the laws of algebra.