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Chapter 9: Problem 63

For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\\{S_{n}\right\\} .\) Then evaluate lim \(S_{n}\) to obtain the value of the series or state that the series diverges. \(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty} \frac{1}{(k+p)(k+p+1)},$$ where \(p\) is a positive integer

Short Answer

Expert verified
The value of the telescoping series is $\frac{1}{1+p}$.

Step by step solution

01

Find the telescoping form

Let's find the telescoping form of the given series. To do this, we'll use partial fractions decomposition. We want to find constants \(A\) and \(B\) such that: $$\frac{1}{(k+p)(k+p+1)} = \frac{A}{k+p} + \frac{B}{k+p+1}.$$ Multiplying both sides by \((k+p)(k+p+1)\), we get: $$1 = A(k+p+1) + B(k+p).$$ Now we can find \(A\) and \(B\). Let \(k = -p\), then: $$1 = A(-p+p+1) \Rightarrow A = 1.$$ Let \(k = -p-1\), then: $$1 = B(-p-1+p) \Rightarrow B = -1.$$ So, the telescoping form of the series is: $$\sum_{k=1}^{\infty} \left(\frac{1}{k+p} - \frac{1}{k+p+1}\right).$$
02

Find the partial sum formula

Next, let's find a formula for the partial sums \(S_n\). We can rewrite the series as the telescoping form: $$S_n = \sum_{k=1}^{n} \left(\frac{1}{k+p} - \frac{1}{k+p+1}\right).$$ Notice that the terms telescope (i.e., most terms cancel out). The partial sum becomes: $$S_n = \left(\frac{1}{1+p} - \frac{1}{2+p}\right) + \left(\frac{1}{2+p} - \frac{1}{3+p}\right) + \cdots + \left(\frac{1}{n+p} - \frac{1}{n+p+1}\right).$$ We can see that most terms cancel out: $$S_n = \frac{1}{1+p} - \frac{1}{n+p+1}.$$
03

Evaluate the limit

Now, we need to evaluate the limit of \(S_n\) as \(n\) approaches infinity: $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(\frac{1}{1+p} - \frac{1}{n+p+1}\right).$$ As \(n\) approaches infinity, the second term approaches 0: $$\lim_{n \to \infty} S_n = \frac{1}{1+p} - 0 = \frac{1}{1+p}.$$ So, the value of the telescoping series is \(\frac{1}{1+p}\).

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Key Concepts

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

Partial Fractions
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler ones. This is especially useful when dealing with integrals or series, as it simplifies the process of finding sums or integrals by making the expression more manageable.
For instance, consider a fraction \( \frac{1}{(k+p)(k+p+1)} \). We can decompose it into simpler fractions:
  • Find constants \(A\) and \(B\) such that: \( \frac{1}{(k+p)(k+p+1)} = \frac{A}{k+p} + \frac{B}{k+p+1} \).
  • This involves setting up the equation: \( 1 = A(k+p+1) + B(k+p) \), and solving for \(A\) and \(B\) by substituting suitable values for \(k\).
  • In our example, these are determined as \(A = 1\) and \(B = -1\).
Once you have these constants, the original expression is represented as a difference: \( \frac{1}{k+p} - \frac{1}{k+p+1} \). This transformation is the essence of the telescoping series.
Partial Sums
The concept of partial sums is central to understanding infinite series. A partial sum, denoted as \(S_n\), is the sum of the first \(n\) terms of a series.
  • For telescoping series, this involves a sequence where most terms cancel each other, simplifying the expression to a few surviving terms.
  • Take the series \( S_n = \sum_{k=1}^{n} \left(\frac{1}{k+p} - \frac{1}{k+p+1}\right). \)
  • Here, each pair of terms like \( \frac{1}{k+p} \) and \(- \frac{1}{k+p}\) cancel, leaving only \( \frac{1}{1+p} - \frac{1}{n+p+1} \) after simplification.
Because of this cancellation, telescoping series require fewer computations, making them easier to manage and understand over standard mathematical manipulations.
Series Convergence
Convergence in series determines if the series reaches a specific value as more terms are added. For our example, the convergence is evaluated by considering the limit of its partial sums as \(n\) approaches infinity.
  • Look at the partial sum \( S_n = \frac{1}{1+p} - \frac{1}{n+p+1} \).
  • As \(n\) becomes very large, \( \frac{1}{n+p+1} \) tends to zero because the denominator grows indefinitely.
  • This means \( \lim_{n \to \infty} S_n = \frac{1}{1+p} \).
This calculation shows that the series converges to \(\frac{1}{1+p}\), providing a single finite value. Knowing this helps in a broad range of applications like physics and engineering where the behavior of infinite processes needs to be understood in manageable terms.

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Most popular questions from this chapter

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}}=\frac{7 \pi^{4}}{720}.\) (Assume the result of Exercise 63.)

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$ \frac{1}{n+1}>\ln (n+2)-\ln (n+1) $$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\) e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\},\) estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured, but not proved, that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

After many nights of observation, you notice that if you oversleep one night you tend to undersleep the following night, and vice versa. This pattern of compensation is described by the relationship $$x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right), \quad \text { for } n=1,2,3, \ldots$$ where \(x_{n}\) is the number of hours of sleep you get on the \(n\) th night and \(x_{0}=7\) and \(x_{1}=6\) are the number of hours of sleep on the first two nights, respectively. a. Write out the first six terms of the sequence \(\left\\{x_{n}\right\\}\) and confirm that the terms alternately increase and decrease. b. Show that the explicit formula $$x_{n}=\frac{19}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{n}, \text { for } n \geq 0,$$ generates the terms of the sequence in part (a). c. What is the limit of the sequence?

Reciprocals of odd squares Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\)

Consider the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}},\) where \(p\) is a real number. a. For what values of \(p\) does this series converge? b. Which of the following series converges faster? Explain. $$ \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { or } \sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}} ? $$
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