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Chapter 11: Problem 2

Find the sum for each of the series: a. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) b. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). c. \(\sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right)\). d. \(\sum_{n=1}^{\infty} \frac{3}{n(n+3)} .\)

Short Answer

Expert verified
The sums for the series are \(\frac{12}{5}\), \(\frac{1}{10}\), \(\frac{23}{2}\), and \(\frac{11}{6}\) respectively.

Step by step solution

01

Solve the first series

The first series is given by \( \sum_{n=0}^{\infty} \frac{(-1)^n 3}{4^n} \). The sum \( S \) can be found using the formula \( \frac{a}{1 - r} \), where \( a = 3 \) and \( r = -\frac{1}{4} \), i.e. \( S = \frac{3}{1 - (-\frac{1}{4})} = \frac{3}{1 + \frac{1}{4}} = \frac{3}{\frac{5}{4}} = \frac{12}{5} \).
02

Solve the second series

The second series is given by \( \sum_{n=2}^{\infty} \frac{2}{5^n} \). We can redefine \( n \) such that the summation starts from 0. So, it becomes \( \sum_{n=0}^{\infty} \frac{2}{5^{n+2}} = \sum_{n=0}^{\infty} \frac{2}{25} \times \frac{1}{5^n} \). This becomes a geometric series with \( a = \frac{2}{25} \) and \( r = \frac{1}{5} \). Using the sum formula, we get \( S = \frac{a}{1 - r} = \frac{2/25}{1 - \frac{1}{5}} = \frac{2/25}{\frac{4}{5}} = \frac{2/25}{\frac{5}{4}} = \frac{2}{20} = \frac{1}{10} \).
03

Solve the third series

The third series is \( \sum_{n=0}^{\infty} (\frac{5}{2^n} + \frac{1}{3^n}) \). Splitting the series, we get \( \sum_{n=0}^{\infty} \frac{5}{2^n} + \sum_{n=0}^{\infty} \frac{1}{3^n} \). Solving individually, we get \( S1 = \frac{5}{1 - \frac{1}{2}} = 10 \) and \( S2 = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \). Summing these up, \( S = S1 + S2 = 10 + \frac{3}{2} = \frac{23}{2} \).
04

Solve the fourth series

The fourth series given is \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} \). This can be rewritten as \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} = \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+3}) \). The series telescopes and thus, the sum is \( S = 1 - \frac{1}{4} + \frac{1}{2} - \frac{1}{5} + \frac{1}{3} - \frac{1}{6} + ... = 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} = \frac{11}{6} \).

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Most popular questions from this chapter

Do the following: a. Compute: \(\lim _{n \rightarrow \infty} n \ln \left(1-\frac{3}{n}\right)\). b. Use L'Hopital's Rule to evaluate \(L=\lim _{x \rightarrow \infty}\left(1-\frac{4}{x}\right)^{x}\). [Hint: Consider \(\ln L\).] c. Determine the convergence of \(\sum_{n=1}^{\infty}\left(\frac{n}{3 n+2}\right)^{n^{2}}\). d. Sum the series \(\sum_{n=1}^{\infty}\left[\tan ^{-1} n-\tan ^{-1}(n+1)\right]\) by first writing the \(N\) th partial sum and then computing \(\lim _{N \rightarrow \infty} s_{N}\).

Consider the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}\) a. Use an appropriate convergence test to show that this series converges. b. Verify that $$ \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) $$ c. Find the \(n\)th partial sum of the series \(\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right)\) and use it to determine the sum of the resulting telescoping series.

Use deMoivre's Theorem to write \(\sin ^{3} \theta\) in terms of \(\sin \theta\) and \(\sin 3 \theta .\)

Determine if the following converge, or diverge, using one of the convergence tests. If the series converges, is it absolute or conditional? a. \(\sum_{n=1}^{\infty} \frac{n+4}{2 n^{3}+1} .\) b. \(\sum_{n=1}^{\infty} \frac{\sin n}{n^{2}}\) c. \(\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^{2}}\). d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n-1}{2 n^{2}-3} .\) e. \(\sum_{n=1}^{\infty} \frac{\ln n}{n}\) f. \(\sum_{n=1}^{\infty} \frac{100^{n}}{n^{200}} .\) g. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n}{n+3}\). h. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{\sqrt{5 n}}{n+1}\).

Determine the radius and interval of convergence of the following infinite series: a. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(x-1)^{n}}{n}\) b. \(\sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} n !}\) c. \(\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{x}{5}\right)^{n}\) d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{x^{n}}{\sqrt{n}}\)
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