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Chapter 9: Problem 48

Choose your test Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{5^{k}-1}$$

Short Answer

Expert verified
Explain. Answer: The series converges. By applying the Ratio Test and computing the limit of the ratio of consecutive terms, we found that the limit is $$\frac{1}{5}<1$$. Since the limit of the ratio is less than 1, the series converges according to the Ratio Test.

Step by step solution

01

Set up the Ratio Test

Let's denote the series terms as \(a_k = \frac{1}{5^k - 1}\). To apply the Ratio Test, we need to calculate the limit of the ratio of consecutive terms, that is, $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|$$
02

Calculate the Ratio

Substitute the definition of \(a_k\) into the ratio. It becomes: $$\lim_{k \to \infty} \frac{a_{k+1}}{a_{k}} = \lim_{k \to \infty} \left|\frac{\frac{1}{5^{k+1}-1}}{\frac{1}{5^k-1}}\right| = \lim_{k \to \infty} \left|\frac{5^k-1}{5^{k+1}-1}\right|$$
03

Simplify the Ratio

Factor out a \(5^k\) from both the numerator and the denominator to simplify the expression: $$\lim_{k \to \infty} \left|\frac{5^k(1-\frac{1}{5^k})}{5^k(5-\frac{1}{5^k})}\right| = \lim_{k \to \infty} \left|\frac{1-\frac{1}{5^k}}{5-\frac{1}{5^k}}\right|$$
04

Calculate the Limit

As \(k\) goes to infinity, the term \(\frac{1}{5^k}\) approaches zero. Thus, the limit becomes: $$\lim_{k \to \infty} \left|\frac{1-\frac{1}{5^k}}{5-\frac{1}{5^k}}\right| = \left|\frac{1}{5}\right|$$
05

Test Convergence

Now that we have found the limit of the ratio, we can apply the Ratio Test to determine the convergence of the series: - The limit is $$\frac{1}{5}<1$$, so the series $$\sum_{k=1}^{\infty} \frac{1}{5^{k}-1}$$ converges.

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Most popular questions from this chapter

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd. } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N,\) the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

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 geometric series $$S=\sum_{k=0}^{\infty} r^{k}$$ which has the value \(1 /(1-r)\) provided \(|r|<1 .\) Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$R_{n}=\left|S-S_{n}\right|=\left|\frac{r^{n}}{1-r}\right|$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \dots$$

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