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Chapter 7: Problem 37

Write the series using summation notation (starting with \(k=1\) ). Each series is either an arithmetic series or a geometric series. $$ \frac{5}{9}+\frac{5}{27}+\frac{5}{81}+\cdots+\frac{5}{3^{40}} $$

Short Answer

Expert verified
The given series can be represented in summation notation as: \[\sum_{k=1}^{40} \frac{5}{3^k}\].

Step by step solution

01

Identify the geometric series

We are given the series: \( \frac{5}{9}+\frac{5}{27}+\frac{5}{81}+\cdots+\frac{5}{3^{40}} \) Each term in the series can be represented as \(\frac{5}{3^k}\) for some positive integer \(k\). Notice that the ratio between consecutive terms is constant: \[\frac{\frac{5}{3^2}}{\frac{5}{3^1}} = \frac{\frac{5}{3^3}}{\frac{5}{3^2}} =\cdots = \frac{\frac{5}{3^{40}}}{\frac{5}{3^{39}}}= \frac{1}{3}\] Since the ratio between consecutive terms is constant, we can conclude that the given series is a geometric series.
02

Find the first term and common ratio

Based on the observation above, we can identify the first term and the common ratio for this geometric series. The first term is: \(a_1 = \frac{5}{3^1}= \frac{5}{3}\) The common ratio is: \(r = \frac{1}{3}\) Now, we have everything we need to write the series using the summation notation.
03

Write the series in summation notation

The given series can be written in summation notation as: \[\sum_{k=1}^{40} a_1 \cdot r^{k-1}\] Substitute the values for \(a_1\) and \(r\): \[\sum_{k=1}^{40} \left(\frac{5}{3}\right) \cdot \left( \frac{1}{3} \right)^{k-1}\] Thus, the series in summation notation is: \[\sum_{k=1}^{40} \frac{5}{3^k}\]

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