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Chapter 3: Problem 6

Show that (a) \(\lim \left(\frac{1}{\sqrt{n+7}}\right)=0\), (b) \(\lim \left(\frac{2 n}{n+2}\right)=2\), (c) \(\lim \left(\frac{\sqrt{n}}{n+1}\right)=0\), (d) \(\lim \left(\frac{(-1)^{n} n}{n^{2}+1}\right)=0\).

Short Answer

Expert verified
The solutions are (a) 0, (b) 2, (c) 0, (d) 0.

Step by step solution

01

(a) Solving \(\lim \left(\frac{1}{\sqrt{n+7}}\right)\)

As n approaches infinity, any fixed number divided by \(\sqrt{n}\) tends towards 0 because the n in the denominator increases faster than the numerator, causing the whole fraction to tend towards 0. Hence: \(\lim \left(\frac{1}{\sqrt{n+7}}\right)=0\)
02

(b) Solving \(\lim \left(\frac{2n}{n+2}\right)\)

To compute this limit, divide both the numerator and the denominator by n (the highest power in the denominator). This gives \(\lim \left(\frac{2}{1+(2/n)}\right)\). As n approaches infinity, the 2/n term becomes zero, this limit simplifies to \(\lim \left(\frac{2}{1+0}\right)\). So, \(\lim \left(\frac{2n}{n+2}\right) = 2\)
03

(c) Solving \(\lim \left(\frac{\sqrt{n}}{n+1}\right)\)

Following a similar process to earlier steps, divide the numerator and denominator by \(\sqrt{n}\) to get \(\lim \left(\frac{1}{\sqrt{n} + 1/\sqrt{n}}\right)\). As n approaches infinity, both terms in the denominator approach zero, so \(\lim \left(\frac{\sqrt{n}}{n+1}\right)=0\)
04

(d) Solving \(\lim \left(\frac{(-1)^n n}{n^2+1}\right)\)

Divide the numerator and denominator by \(n^2\) to get \(\lim \left(\frac{(-1)^n /n}{1 + 1/n^2}\right)\). As n approaches infinity, \(-1^n/n\) approaches 0 and \(1/n^2\) also approaches zero. Hence, the limit is zero.

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Most popular questions from this chapter

Let \(\sum a_{n}\) be a given series and let \(\sum b_{n}\) be the series in which the terms are the same and in the same order as in \(\sum a_{n}\) except that the terms for which \(a_{n}=0\) have been omitted. Show that \(\sum a_{n}\) converges to \(A\) if and only if \(\sum b_{n}\) converges to \(A\).

If \(x_{1}>0\) and \(x_{n+1}:=\left(2+x_{n}\right)^{-1}\) for \(n \geq 1\), show that \(\left(x_{n}\right)\) is a contractive sequence. Find the limit.

Show that \(\lim \left(\frac{1}{n}-\frac{1}{n+1}\right)=0\).

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n} a_{n+1}}\) always convergent? Either prove it or give a counterexample.

Give an example of two divergent sequences \(X\) and \(Y\) such that: (a) their sum \(X+Y\) converges, (b) their product \(X Y\) converges.
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