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Chapter 9: Problem 50

Use the properties of infinite series to evaluate the following series. $$\sum_{k=0}^{\infty} \frac{2-3^{k}}{6^{k}}$$

Short Answer

Expert verified
Answer: The sum of the given infinite series is 1.

Step by step solution

01

Split the series into two geometric series

We start by splitting the given series into two separate geometric series: $$\sum_{k=0}^{\infty} \frac{2}{6^{k}} - \sum_{k=0}^{\infty} \frac{3^{k}}{6^{k}}$$
02

Simplify the series expressions

Now, we simplify each of the geometric series. For the first geometric series, it remains the same. For the second geometric series, we can simplify it as follows: $$\frac{3^{k}}{6^{k}} = \frac{3^{k}}{3^{k} \cdot 2^{k}} = \frac{1}{2^{k}}$$ So, our separated series now become: $$\sum_{k=0}^{\infty} \frac{2}{6^{k}} - \sum_{k=0}^{\infty} \frac{1}{2^{k}}$$
03

Evaluate the geometric series sums

As both separated series are geometric series, we can now evaluate their sums using the geometric series sum formula: $$S = \frac{a}{1-r}$$ Where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio. For the first series, \(a = \frac{2}{6^0} = 2\) and \(r = \frac{1}{3}\). Therefore, $$S_1 = \frac{2}{1-\frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3$$ For the second series, \(a = \frac{1}{2^0} = 1\) and \(r = \frac{1}{2}\). Therefore, $$S_2 = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$$
04

Find the overall sum

Now that we have the sums of the two separate geometric series, we just need to combine them to find the overall sum of the given series. Since we have subtraction between the two series, we subtract the sums: $$S = S_1 - S_2 = 3 - 2 = 1$$ So, the sum of the given infinite series is: $$\sum_{k=0}^{\infty} \frac{2-3^{k}}{6^{k}} = 1$$

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Most popular questions from this chapter

Two sine series Determine whether the following series converge. a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\)

Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$ \cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}}$$

For a positive real number \(p,\) how do you interpret \(p^{p^{p \cdot *}},\) where the tower of exponents continues indefinitely? As it stands, the expression is ambiguous. The tower could be built from the top or from the bottom; that is, it could be evaluated by the recurrence relations \(a_{n+1}=p^{a_{n}}\) (building from the bottom) or \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{0}=p\) in either case. The two recurrence relations have very different behaviors that depend on the value of \(p\). a. Use computations with various values of \(p > 0\) to find the values of \(p\) such that the sequence defined by (2) has a limit. Estimate the maximum value of \(p\) for which the sequence has a limit. b. Show that the sequence defined by (1) has a limit for certain values of \(p\). Make a table showing the approximate value of the tower for various values of \(p .\) Estimate the maximum value of \(p\) for which the sequence has a limit.

Evaluate the limit of the following sequences. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$
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