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

Short Answer

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Question: Determine the value of the given infinite series: $$\sum_{k=2}^{\infty} 3 e^{-k}.$$ Answer: The value of the given series is $$ S = \frac{3e^{-1}}{e - 1}.$$

Step by step solution

01

Identify the first term and the common ratio

The given series is: $$\sum_{k=2}^{\infty} 3 e^{-k}$$ In this series, the first term occurs when \(k = 2\), which is: $$a = 3e^{-2}$$ The common ratio can be identified as $$r = e^{-1}$$.
02

Verify the convergence of the series

Since the series is geometric and \(|r| = e^{-1} < 1\), it converges.
03

Apply the sum formula

We know that the sum of an infinite geometric series can be found using the formula: $$ S = \frac{a}{1 - r}$$ Now, substitute the values of \(a\) and \(r\) in the formula: $$ S = \frac{3e^{-2}}{1 - e^{-1}}$$
04

Simplify the expression

Simplify the expression to get the final value of the series: $$ S = \frac{3 e^{-2}}{1 - e^{-1}}$$ $$ S = \frac{3 e^{-2}}{e^{-1}(e - 1)}$$ $$ S = \frac{3e^{-1}}{e - 1}$$ The value of the given series is: $$ S = \frac{3e^{-1}}{e - 1}$$

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Most popular questions from this chapter

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

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