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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

Compute the following integrals: a. \(\int x e^{2 x^{2}} d x\) b. \(\int_{0}^{3} \frac{5 x}{\sqrt{x^{2}+16}} d x\) c. \(\int x^{3} \sin 3 x d x\). (Do this using integration by parts, the Tabular Method, and differentiation under the integral sign.) d. \(\int \cos ^{4} 3 x d x\) e. \(\int_{0}^{\pi / 4} \sec ^{3} x d x\) f. \(\int e^{x} \sinh x d x\) g. \(\int \sqrt{9-x^{2}} d x\) h. \(\int \frac{d x}{\left(4-x^{2}\right)^{2}}\), using the substitution \(x=2 \tanh u\). i. \(\int_{0}^{4} \frac{d x}{\sqrt{9+x^{2}}}\), using a hyperbolic function substitution. j. \(\int \frac{d x}{1-x^{2}}\), using the substitution \(x=\tanh u\). k. \(\int \frac{d x}{\left(x^{2}+4\right)^{3 / 2}}\), using the substitutions \(x=2 \tan \theta\) and \(x=2 \sinh u\). 1\. \(\int \frac{d x}{\sqrt{3 x^{2}-6 x+4}}\)

Write the following in terms of logarithms: a. \(\cosh ^{-1} \frac{4}{3}\). b. \(\tanh ^{-1} \frac{1}{2}\). c. \(\sinh ^{-1} 2\).

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 \(0.569 ?\)

Determine the exact values of a. \(\sin \frac{\pi}{8}\). b. \(\tan 15^{\circ}\). c. \(\cos 105^{\circ}\).

Evaluate the following expressions at the given point. Use your calculator or your computer (such as Maple). Then use series expansions to find an approximation of the value of the expression to as many places as you trust. a. \(\frac{1}{\sqrt{1+x^{3}}}-\cos x^{2}\) at \(x=0.015\). b. \(\ln \sqrt{\frac{1+x}{1-x}}-\tan x\) at \(x=0.0015\). c. \(f(x)=\frac{1}{\sqrt{1+2 x^{2}}}-1+x^{2}\) at \(x=5.00 \times 10^{-3}\). d. \(f(R, h)=R-\sqrt{R^{2}+h^{2}}\) for \(R=1.374 \times 10^{3} \mathrm{~km}\) and \(h=1.00 \mathrm{~m}\). e. \(f(x)=1-\frac{1}{\sqrt{1-x}}\) for \(x=2.5 \times 10^{-13}\).
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