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Chapter 10: Problem 35

In Exercises \(35-40,\) find a formula for the \(n\) th partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) $$

Short Answer

Expert verified
The series converges, and its sum is 1.

Step by step solution

01

Understanding the Series

The given series is \( \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) \). Each term in the series is of the form \( a_n = \frac{1}{n} - \frac{1}{n+1} \). This type of series is known as a telescoping series because consecutive terms cancel each other out in a pattern.
02

Recognize the Telescoping Nature

To see the telescoping nature, compute the first few terms:\[a_1 = \frac{1}{1} - \frac{1}{2}\], \[a_2 = \frac{1}{2} - \frac{1}{3}\], \[a_3 = \frac{1}{3} - \frac{1}{4}\]. Notice that in the sum \((a_1 + a_2 + a_3)\), most intermediate terms cancel out, leaving only \(\frac{1}{1} - \frac{1}{4}\).
03

Find the n-th Partial Sum

The n-th partial sum \(S_n\) of the series is\[S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right).\] This simplifies to \[S_n = 1 - \frac{1}{n+1}.\]
04

Determine Convergence of the Series

To determine if the series converges, evaluate the limit of the n-th partial sum as \(n \to \infty\):\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1.\] Since this limit exists and is finite, the series converges.
05

Conclude the Sum of the Series

Since the series converges, and the limit of the n-th partial sum \(S_n\) as \(n\) approaches infinity is 1, the sum of the series is 1.

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Key Concepts

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

series convergence
In mathematical series analysis, understanding when a series converges is crucial. Series convergence refers to a condition where the sum of the series approaches a finite value as the number of terms increases to infinity. For a series to converge, it must continuously add up to a specific value rather than growing indefinitely.
For instance, consider the series:
  • The series is expressed as an infinite sum, \( \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) \).
  • Convergence in this context means checking if the limit of its n-th partial sum exists.
In simpler terms, you look for what value the series adds up to as you include more and more terms. If the series produces a finite sum, it is convergent, as is the case in our exercise where the series converges to 1.
partial sum formula
A partial sum formula helps us find the sum of the first \( n \) terms of a series. In a telescoping series, like the one in our exercise, consecutive terms cancel out.
This can make finding the partial sum easier:
  • The given series is: \( \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) \).
  • Each term \( a_n = \frac{1}{n} - \frac{1}{n+1} \) results in cancellation between successive terms.
  • The n-th partial sum, \( S_n \), simplifies greatly due to this characteristic.
After canceling the intermediate terms, the formula comes down to: \[ S_n = 1 - \frac{1}{n+1} \]. This shows how telescoping series can lead to simple partial sum formulas, making it easy to compute the behavior of the series as a whole.
mathematical series analysis
Mathematical series analysis involves investigating the behavior of series, focusing on convergence, divergence, and sum computations. The series given in the exercise is a typical example of a telescoping series where the main tool is the concept of partial sums.
Key aspects include:
  • Identifying series types, like telescoping, where terms cancel each other.
  • Using partial sum formulas to compute the sum of a finite number of terms, simplifying more intricate calculations.
  • Analyzing the limit of partial sums will determine convergence.
For the exercise, by analyzing the n-th partial sum, \( S_n = 1 - \frac{1}{n+1} \), and evaluating its limit as \( n \to \infty \), it becomes clear that the series converges.
With a final result showing the series converges to 1, series analysis provides a structured approach to solving complex mathematical problems involving infinite sums.

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Most popular questions from this chapter

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\operatorname{coth} n}{n^{2}}\end{equation}

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Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)
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