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Chapter 8: Problem 55

Find all values of \(p\) such that the sequence \(a_{n}=\frac{1}{p^{n}}\) converges.

Short Answer

Expert verified
The sequence \(a_{n}=\frac{1}{p^{n}}\) converges for all \(p > 1\).

Step by step solution

01

Analyse the given sequence

First, inspect the sequence \(a_{n}=\frac{1}{p^{n}}\). Upon close examination, it can be seen that if \(p\) is positive, and bigger than 1, then the denominator grows faster than the numerator as \(n\) increases. This means the terms of the sequence will get closer and closer to 0 as \(n\) increases. On the other hand, if \(p\) is positive but less than or equal to 1, then the numerator grows faster than the denominator as \(n\) increases. This means the terms of the sequence will not get closer to any specific value as \(n\) goes to infinity i.e the sequence diverges.
02

Formulate the condition for convergence

From the analysis in step 1, it can be concluded that for the sequence \(a_{n}\) to converge, \(p\) must be positive and greater than 1. Mathematically this can be written as \(p>1\).
03

Summarise the result

The deduction from the above steps concludes the solution. For the sequence \(a_{n}=\frac{1}{p^{n}}\) to converge, \(p\) must be a positive number greater than 1. If it is not, the sequence will diverge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergent Sequences
A convergent sequence is a sequence in which the terms approach a specific value, known as a limit, as the sequence progresses towards infinity. In simpler terms, as you progress further along the sequence, the values get closer and closer to this single number.
  • This behavior provides us with predictability, where we can guess where the sequence is headed.
  • For instance, the sequence given in the exercise, \(a_{n} = \frac{1}{p^{n}}\), becomes very small and approaches zero if \(p > 1\).
In general, for convergence:
  • As \(n\) (the term position) becomes a very large number, if the sequence approaches a finite number, it's considered convergent.
  • If it doesn't settle toward any particular number, then the sequence diverges.
Infinite Sequences
Infinite sequences are those that continue indefinitely, with no final term. This means they go on forever, which is important when considering convergence and divergence.
  • The sequence \(a_{n} = \frac{1}{p^{n}}\) from our exercise is an infinite sequence, as we can keep increasing \(n\) infinitely.
  • This infinite nature means we can keep testing if it heads towards a specific number as \(n\) keeps increasing.
Understanding infinite sequences helps us better analyze the long-term behavior of sequences to determine if they converge or diverge, based on the values taken by the sequence as \(n\) approaches infinity.
Positive Numbers
Positive numbers are numbers greater than zero. In the context of the exercise, the positivity of \(p\) is crucial.
  • For convergence, \(p\) must be a positive number greater than 1. This ensures the denominator in \(\frac{1}{p^{n}}\) grows rapidly, causing the terms to shrink.
  • If \(p\) were not positive (or even just equal to 1), the shrinking effect wouldn鈥檛 occur the same way.
Therefore, positivity ensures the components of the sequence behave predictably, influencing whether the sequence converges or not.

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Most popular questions from this chapter

Determine the radius and interval of convergence. $$\sum_{k=0}^{\infty} k !(x+1)^{k}$$

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

Use the even/odd properties of \(f(x)\) to predict (don't compute) whether the Fourier series will contain only cosine terms, only sine terms or both. $$f(x)=|x|$$

Find the Taylor series for \(\sqrt{x}\) about a general center \(c=a^{2}\)

(a) use a Taylor polynomial of degree 4 to approximate the given number, (b) estimate the error in the approximation and (c) estimate the number of terms needed in a Taylor polynomial to guarantee an accuracy of \(10^{-10}\) $$\ln (1.05)$$
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