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Chapter 3: Problem 11

If \(0

Short Answer

Expert verified
The limit \(\lim \left(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\right)\) as \(n\) goes to infinity is 1.

Step by step solution

01

Expression of the Ratio

The given expression is \(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\). This can be rewritten as \(\frac{a*a^{n}+b*b^{n}}{a^{n}+b^{n}}\) which is allowed since \(a^{n+1}=a*a^{n}\) and \(b^{n+1}=b*b^{n}\).
02

Use of properties of Limit

The limit law that the limit of a sum of two functions is the sum of their limits can now be used, hence we can rewrite \(\lim \left(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\right)\) as \(\lim \left(\frac{a*a^{n}}{a^{n}+b^{n}}\right) + \lim \left(\frac{b*b^{n}}{a^{n}+b^{n}}\right) \). Due to the given condition \(0<a<b\), we observe that \(\frac{a*a^{n}}{a^{n}+b^{n}}\) tends towards 0 while \(\frac{b*b^{n}}{a^{n}+b^{n}}\) tends towards 1.
03

Observation of Trends

By observing the limit of the two seperated fractions, it can be noticed that the first component \(\frac{a*a^{n}}{a^{n}+b^{n}}\) goes to zero because the numerator \(a*a^{n}\) is always going to be less than \(a^{n}+b^{n}\) as \(a<b\). Therefore, as \(n\) goes to infinity, this fraction goes to 0. For the second term \(\frac{b*b^{n}}{a^{n}+b^{n}}\), since \(b\) is greater than \(a\), as \(n\) goes to infinity, the \(b*b^{n}\) term will be significantly larger than \(a^{n}\) and therefore, this fraction will go towards 1 as \(n\) goes to infinity.

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Most popular questions from this chapter

Show directly from the definition that the following are Cauchy sequences. (a) \(\left(\frac{n+1}{n}\right)\). (b) \(\left(1+\frac{1}{2 !}+\cdots+\frac{1}{n !}\right)\).

Use the definition of the limit of a sequence to establish the following limits. (a) \(\lim \left(\frac{n}{n^{2}+1}\right)=0\), (b) \(\lim \left(\frac{2 n}{n+1}\right)=2\), (c) \(\lim \left(\frac{3 n+1}{2 n+5}\right)=\frac{3}{2}\), (d) \(\lim \left(\frac{n^{2}-1}{2 n^{2}+3}\right)=\frac{1}{2}\).

Let \(X=\left(x_{n}\right)\) be a sequence of positive real numbers such that \(\lim \left(x_{n+1} / x_{n}\right)=L>1 .\) Show that \(X\) is not a bounded sequence and hence is not convergent.

Show directly from the definition that the following are not Cauchy sequences. (a) \(\left((-1)^{n}\right)\), (b) \(\left(n+\frac{(-1)^{n}}{n}\right)\), (c) \((\ln n)\).

If \(x_{1}2\), show that \(\left(x_{n}\right)\) is convergent. What is its limit?
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