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Chapter 11: Problem 5

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.

Short Answer

Expert verified
The given series converges with a sum of \(1/2\).

Step by step solution

01

Apply Convergence Test

To analyze the convergence of the series, one can apply the Ratio Test, having that the terms are all positive. Therefore, a ratio between \(a_{n+1}/a_n\) must be calculated where \(a_n\) is the nth term of the series. This ratio tends to 1 as \(n\) approaches infinity, which means the series converges due to the ratio test.
02

Verify the Series Equality

In this step, the series equality is verified by performing a fractional decomposition of \(\frac{1}{(n+2)(n+1)}\), which results in \(frac{n+1}{n+2}-\frac{n}{n+1}\) as given. This confirms the equality.
03

Find the Nth Partial Sum

The nth partial sum of the series can be computed by summing up the first ‘n’ terms of the series. Since this is a telescoping series, most terms cancel out, and the partial sum \(\frac{n+1}{n+2}-\frac{1}{2}\) remains.
04

Determine the Sum of the Telescoping Series

The sum of the telescoping series can be calculated by taking the limit of the nth partial sum as \(n\) approaches infinity, which results in \(1-1/2 = 1/2\).

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Most popular questions from this chapter

Evaluate the following expressions at the given point. Use your calculator and your computer (such as Maple). Then use series expansions to find an approximation to 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 \(x=2.5 \times 10^{-13}\)

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. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{\sqrt{5 n}}{n+1}\).

Recall that the alternating harmonic series converges conditionally. a. From the Taylor series expansion for \(f(x)=\ln (1+x)\), inserting \(x=1\) gives the alternating harmonic series. What is the sum of the alternating harmonic series? b Because the alternating harmonic series does not converge absolutely, a rearrangement of the terms in the series will result in series whose sums vary. One such rearrangement in alternating \(p\) positive terms and \(n\) negative terms leads to the following sum \(^{10}\) : $$ \begin{gathered} \frac{1}{2} \ln \frac{4 p}{n}=\underbrace{\left(1+\frac{1}{3}+\cdots+\frac{1}{2 p-1}\right)}_{p \text { terms }}-\underbrace{\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)}_{n \text { terms }} \\ +\underbrace{\left(\frac{1}{2 p+1}+\cdots+\frac{1}{4 p-1}\right)}_{p \text { terms }}-\underbrace{\left(\frac{1}{2 n+2}+\cdots+\frac{1}{4 n}\right)}_{n \text { terms }}+\cdots . \end{gathered} $$ Find rearrangements of the alternating harmonic series to give the following sums; that is, determine \(p\) and \(n\) for the given expression and write down the above series explicitly; that is, determine \(p\) and \(n\) leading to the following sums. i. \(\frac{5}{2} \ln 2\). ii. \(\ln 8\). iii. \(0 .\) iv. A sum that is close to \(\pi\).

Determine the order, \(O\left(x^{p}\right)\), of the following functions. You may need to use series expansions in powers of \(x\) when \(x \rightarrow 0\), or series expansions in powers of \(1 / x\) when \(x \rightarrow \infty\) a. \(\sqrt{x(1-x)}\) as \(x \rightarrow 0\). b. \(\frac{x^{5 / 4}}{1-\cos x}\) as \(x \rightarrow 0\) c. \(\frac{x}{x^{2}-1}\) as \(x \rightarrow \infty\). d. \(\sqrt{x^{2}+x}-x\) as \(x \rightarrow \infty\).
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