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Chapter 9: Problem 72

Identify the two series that are the same. (a) \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{(n-1) 2^{n-1}}\) (b) \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n 2^{n}}\) (c) \(\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(n+1) 2^{n}}\)

Short Answer

Expert verified
The two series that are the same are series (a) and series (b).

Step by step solution

01

Index Adjustments

Note that the indices in the given series start from different values. Adjust each series so that they all start at the same index, the index used will be n = 1, which is the most common in scenarios like this. To do this, replace each \(n\) by \(n + k\), where \(k\) is the desired shift (this would be the difference between the current starting index and the desired starting index).
02

Series (a) Analysis

Having n starting at 2, apply the index shift for series (a), for series (a), \(k = 1\). This transforms the series (a) to \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n 2^n}\)
03

Series (b) Analysis

No adjustments needed for Series (b). It remains \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n 2^n}\)
04

Series (c) Analysis

For series (c), having n starting at 0, apply the index shift. Here, \(k = 1\). This will transform the series (c) into \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1) 2^n}\)
05

Comparison of Series

Compare the standardised series (a), (b), and (c). (a) and (b) are now identical, but (c) is different from both.

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Most popular questions from this chapter

Investigation Consider the function \(f\) defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}},} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right.$$ (a) Sketch a graph of the function. (b) Use the alternative form of the definition of the derivative (Section 2.1) and L'H贸pital's Rule to show that \(f^{\prime}(0)=0\) . [By continuing this process, it can be shown that \(f^{(n)}(0)=0\) for \(n>1 . ]\) (c) Using the result in part (b), find the Maclaurin series for \(f\) . Does the series converge to \(f ?\)

Probability In Exercises 73 and \(74,\) approximate the normal probability with an error of less than 0.0001 , where the probability is given by $$P(a < x < b)=\frac{1}{\sqrt{2 \pi}} \int_{a}^{b} e^{-x^{2} / 2} d x$$ $$ P(1 < x < 2) $$

Verify that the Ratio Test is inconclusive for the \(p\) -series. $$\sum_{n=1}^{\infty} \frac{1}{n^{1 / 2}}$$

Use the Ratio Test or the Root Test to determine the convergence or divergence of the series. \(\begin{aligned} 1 &+\frac{1 \cdot 3}{1 \cdot 2 \cdot 3}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} \\ &+\frac{1 \cdot 3 \cdot 5 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}+\cdots \end{aligned}\)

Finding a Limit In Exercises \(59-62,\) use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{1-\cos x}{x} $$
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