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Chapter 9: Problem 14

Show that the series \(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}++-\cdots\) is divergent.

Short Answer

Expert verified
The series \(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}++-\cdots\) is divergent. None of the convergence tests applicable suggest convergence, and the terms do not decrease to zero strictly, therefore it is concluded that the series diverges.

Step by step solution

01

Determine the general term

First examine the series and understand the pattern. The terms alternate between positive and negative, but every two positive terms are followed by a negative term. Express this as a general term. Identify the pattern of positive and negative terms: these follow the sequence \(a_n = (-1)^k / (2n-1)\) for \(k= n + \lfloor \frac{n}{2} \rfloor\) where \(\lfloor x \rfloor\) is the floor function, which rounds down to the nearest integer, and \(n\) is the term number.
02

Evaluate the Limit of the Sequence

The divergence of a series can be proven by showing that the limit of its sequence of terms isn't zero. Compute the limit as \(n\) goes to infinity for the sequence \(a_n\). This is found to be \(0\), which doesn't mean that the series is convergent, since the terms do not strictly decrease.
03

Check The Alternating Series Test

The series looks like an alternating series, where terms are alternating in sign. However, the Alternating Series Test needs the terms \(a_n\) to strictly decrease to \(0\) as \(n\) goes to infinity. Here, the terms \(1/(2n-1)\) increase after every two steps then decrease, hence the test fails.
04

Conclude the divergence

Since the Alternating Series Test failed due to the terms not strictly decreasing, and no other convergence test can be applied, we conclude that the series is divergent.

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Most popular questions from this chapter

Suppose that none of the numbers \(a, b, c\) is a negative integer or zero. Prove that the hypergeometric series $$\frac{a b}{1 ! c}+\frac{a(a+1) b(b+1)}{2 ! c(c+1)}+\frac{a(a+1)(a+2) b(b+1)(b+2)}{3 ! c(c+1)(c+2)}+\cdots$$ is absolutely convergent for \(c>a+b\) and divergent for \(c

Show that if \(|x|<1\), then Arctan \(x=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1} x^{2 n+1}\).

If the partial sums of \(\sum a_{n}\) are bounded, show that the series \(\sum_{n=1}^{\infty} a_{n} e^{-n t}\) converges for \(t>0\).

(a) If \(\sum a_{n}\) is absolutely convergent and \(\left(b_{n}\right)\) is a bounded sequence, show that \(\sum a_{n} b_{n}\) is absolutely convergent. (b) Give an example to show that if the convergence of \(\sum a_{n}\) is conditional and \(\left(b_{n}\right)\) is a bounded sequence, then \(\sum a_{n} b_{n}\) may diverge.

Show by integrating the series for \(1 /(1+x)\) that if \(|x|<1\), then $$\ln (1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^{n}$$
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