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Chapter 1: Problem 17

In the event that a series converges uniformly, one can consider the derivative of the series to arrive at the summation of other infinite series. a. Differentiate the series representation for \(f(x)=\frac{1}{1-x}\) to sum the series \(\sum_{n=1}^{\infty} n x^{n},|x|<1\) b. Use the result from part a to sum the series \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\) c. Sum the series \(\sum_{n=2}^{\infty} n(n-1) x^{n},|x|<1\) d. Use the result from part \(c\) to sum the series \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) e. Use the results from this problem to sum the series \(\sum_{n=4}^{\infty} \frac{n^{2}}{5^{n}}\)

Short Answer

Expert verified
a) The sum of the series is \(\frac{1}{(1-x)^2}\). b) The sum is 25. c) The sum is \(\frac{2} {(1-x)^3}\). d) The sum is 200. e) The sum is 125.

Step by step solution

01

Differentiating the Function

The function \(f(x)=\frac{1}{1-x}\) can be represented as a series, \(\sum_{n=0}^{\infty} x^{n}\). If we differentiate this series term by term, the derived series is \(\sum_{n=1}^{\infty} n x^{n-1}\).
02

Summing the Derived Series

The series \(\sum_{n=1}^{\infty} n x^{n-1}\) can be written as \(\sum_{n=1}^{\infty} n x^{n}\), where \(|x|<1\). Thus, the sum of this series is \(\frac{1}{(1-x)^2}\).
03

Sum a Specific Series

For the series \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\), we substitute \(x=\frac{1}{5}\) into \(\frac{1}{(1-x)^2}\) to get \(\frac{1}{(1-\frac{1}{5})^2} = 25\).
04

Differentiating and Summing another Series

We apply the derivative a second time on the series from Step 1, resulting \(\sum_{n=2}^{\infty} n(n-1)x^{n-2}\). Rearranging, we get \(\sum_{n=2}^{\infty} n(n-1)x^{n}\), which sums up to \(\frac{2} {(1-x)^3}\), where \(|x|<1\).
05

Sum the Derived Series

We take the series \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) and put \(x=\frac{1}{5}\) into \(\frac{2} {(1-x)^3}\), yielding \(200\).
06

Sum the final series

To find the summation of the series \(\sum_{n=4}^{\infty} \frac{n^{2}}{5^{n}}\), subtract 3 times the summation of \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\) (obtained in Step 3) from the summation of \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) (obtained in Step 5). So, \(200 - 3*25 = 125\).

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Most popular questions from this chapter

Find the sum for each of the series: a. \(5+\frac{25}{7}+\frac{125}{49}+\frac{625}{343}+\cdots\) b. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) c. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). d. \(\sum_{n=-1}^{\infty}(-1)^{n+1}\left(\frac{e}{\pi}\right)^{n}\). e. \(\sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right)\). f. \(\sum_{n=1}^{\infty} \frac{3}{n(n+3)}\) g. What is \(0.569 ?\)

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