Chapter 8: Problem 2
Suppose the sequence \(\left\\{a_{n}\right\\}\) is defined by the explicit formula \(a_{n}=1 / n,\) for \(n=1,2,3, \ldots .\) Write out the first five terms of the sequence.
Chapter 8: Problem 2
Suppose the sequence \(\left\\{a_{n}\right\\}\) is defined by the explicit formula \(a_{n}=1 / n,\) for \(n=1,2,3, \ldots .\) Write out the first five terms of the sequence.
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Get started for freeEvaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}$$
The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.
Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$
It can be proved that if a series converges absolutely, then its terms may be summed in any order without changing the value of the series. However, if a series converges conditionally, then the value of the series depends on the order of summation. For example, the (conditionally convergent) alternating harmonic series has the value $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2 $$ Show that by rearranging the terms (so the sign pattern is \(++-\) ), $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2} \ln 2 $$
Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1$$
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