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Chapter 9: Problem 22

Determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{\cos \pi k}{k^{2}}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify that the series is alternating

Notice that the series has the form $$\sum_{k=1}^{\infty} \frac{\cos \pi k}{k^{2}}$$ The cosine function takes on positive and negative values in an alternating way for integer multiples of \(\pi\), so we can rewrite the series as $$\sum_{k=1}^{\infty} (-1)^{k}\frac{1}{k^{2}}$$ Now, we can recognize this as an alternating series where \(a_k = \frac{1}{k^{2}}\).
02

Apply the Alternating Series Test

According to the alternating series test, a series $$\sum_{k=1}^{\infty} (-1)^k a_k$$ converges if 1. The sequence \((a_k)\) is decreasing (\((a_{k+1} \leq a_{k}\)) for all \(k\)), and 2. The limit of \(a_k\) as \(k\) goes to infinity is \(0\). Let's check these conditions for our series.
03

Check if the sequence is decreasing

We have \(a_k = \frac{1}{k^2}\). For \(k \geq 1\), observe that as \(k\) increases, the denominator \(k^2\) becomes larger, making the fractions smaller. Therefore, \(a_{k+1} \leq a_k\) for all \(k \geq 1\). So, the sequence \((a_k)\) is decreasing.
04

Check if the limit of the sequence is 0

Now, let's check if the limit of \(a_k\) as \(k\) goes to infinity is \(0\). $$\lim_{k\rightarrow\infty} \frac{1}{k^2} = 0$$ As \(k\) approaches infinity, the denominator, \(k^2\), grows without bound, making the fraction approach \(0\). Thus, the limit of \(a_k\) as \(k\) goes to infinity is indeed \(0\).
05

Conclude the convergence of the series

Since the sequence \((a_k)\) is decreasing and its limit is \(0\), we can conclude that the alternating series $$\sum_{k=1}^{\infty} (-1)^{k}\frac{1}{k^{2}}$$ converges by the Alternating Series Test, which means that the given series $$\sum_{k=1}^{\infty} \frac{\cos \pi k}{k^{2}}$$ also converges.

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Most popular questions from this chapter

The expression where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of where \(a\) and \(b\) are positive real numbers.

The CORDIC (COordinate Rotation DIgital Calculation) algorithm is used by most calculators to evaluate trigonometric and logarithmic functions. An important number in the CORDIC algorithm, called the aggregate constant, is \(\prod_{n=0}^{\infty} \frac{2^{n}}{\sqrt{1+2^{2 n}}},\) where \(\prod_{n=0}^{N} a_{n}\) represents the product \(a_{0} \cdot a_{1} \cdots a_{N}\). This infinite product is the limit of the sequence $$\left\\{\prod_{n=0}^{0} \frac{2^{n}}{\sqrt{1+2^{2 n}}} \cdot \prod_{n=0}^{1} \frac{2^{n}}{\sqrt{1+2^{2 n}}}, \prod_{n=0}^{2} \frac{2^{n}}{\sqrt{1+2^{2 n}}} \ldots .\right\\}.$$ Estimate the value of the aggregate constant.

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} \cos (\pi k)$$

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).
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