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Chapter 11: Problem 1

For those sequences that converge, find the limit \(\lim _{n \rightarrow \infty} a_{n}\) a. \(a_{n}=\frac{n^{2}+1}{n^{3}+1} .\) b. \(a_{n}=\frac{3 n+1}{n+2}\). c. \(a_{n}=\left(\frac{3}{n}\right)^{1 / n}\). d. \(a_{n}=\frac{2 n^{2}+4 n^{3}}{n^{3}+5 \sqrt{2+n^{6}}} .\) e. \(a_{n}=n \ln \left(1+\frac{1}{n}\right)\). f. \(a_{n}=n \sin \left(\frac{1}{n}\right)\). g. \(a_{n}=\frac{(2 n+3) !}{(n+1) !} .\)

Short Answer

Expert verified
The limits are a) 0, b) 3, c) \(e^{-1}\), d) 4, e) 1, f) 1. The sequence g does not have a finite limit.

Step by step solution

01

Sequence a

The first sequence is \(a_{n}=\frac{n^{2}+1}{n^{3}+1}\). Using L'Hopital's rule to solve this indeterminate form \(\frac{0}{0}\) we obtain \(\lim_{n \rightarrow \infty} \frac{2n}{3n^2} = 0\).
02

Sequence b

The second sequence is \(a_{n}=\frac{3n+1}{n+2}\). Again, using L'Hopital's rule for the form \(\infty/\infty\) we obtain \(\lim_{n \rightarrow \infty} \frac{3}{1} = 3\).
03

Sequence c

The third sequence is \(a_{n}=\left(\frac{3}{n}\right)^{1/n}\). Take the natural log of the sequence to simplify, then apply L'Hopital's rule to obtain that it converges to \(e^{-1}\)
04

Sequence d

The fourth sequence is \(a_{n}=\frac{2n^2+4n^3}{n^3+5\sqrt{2+n^6}}\). By dividing the numerator and the denominator by \(n^3\) we get the limit to be 4.
05

Sequence e

The fifth sequence is \(a_{n}=n\ln(1+\frac{1}{n})\). By replacing \(\frac{1}{n}\) with \(x\), we can simplify the sequence to the form that the limit of is computed to be 1.
06

Sequence f

The sixth sequence is \(a_{n}=n\sin(\frac{1}{n})\). If we replace \(\frac{1}{n}\) with \(x\), this simplifies to \(x\sin(x)\), as \(x\) approaches 0. The limit is 1.
07

Sequence g

The seventh sequence is \(a_{n} = \frac{(2n+3)!}{(n+1)!}\) which clearly goes to \(\infty\) as \(n\) increases, therefore it doesn't have a finite limit.

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Most popular questions from this chapter

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

Test for pointwise and uniform convergence on the given set. [The Weierstraß M-Test might be helpful.] a. \(f(x)=\sum_{n=1}^{\infty} \frac{\ln n x}{n^{2}}, x \in[1,2]\) b. \(f(x)=\sum_{n=1}^{\infty} \frac{1}{3^{n}} \cos \frac{x}{2^{n}}\) on \(R\).

Find the Taylor series centered at \(x=a\) and its corresponding radius of convergence for the given function. In most cases, you need not employ the direct method of computation of the Taylor coefficients. a. \(f(x)=\sinh x, a=0\). b. \(f(x)=\sqrt{1+x}, a=0\). c. \(f(x)=x e^{x}, a=1\). d. \(f(x)=\frac{x-1}{2+x}, a=1\).

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 \(x=2.5 \times 10^{-13}\)
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