/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Find the sum of each series. $... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 44

Find the sum of each series. $$\sum_{n=1}^{\infty} \frac{2 n+1}{n^{2}(n+1)^{2}}$$

Short Answer

Expert verified
The sum of the series is 1.

Step by step solution

01

Break Down the General Term

First, we'll consider each term in the sequence given by \( a_n = \frac{2n+1}{n^2(n+1)^2} \). To find its sum, we need to simplify or transform it into a more manageable form, such as a telescoping series.
02

Partial Fraction Decomposition

To simplify \( a_n \), employ partial fraction decomposition to express it in the form \( \frac{2n+1}{n^2(n+1)^2} = \frac{A}{n} + \frac{B}{n^2} + \frac{C}{n+1} + \frac{D}{(n+1)^2} \). Solve for constants \( A \), \( B \), \( C \), and \( D \) by equating coefficients.
03

Solve for Constants

Multiplying throughout by \( n^2(n+1)^2 \) gives:\[ 2n + 1 = A n(n+1)^2 + B (n+1)^2 + C n^2(n+1) + D n^2 \]Expanding and equating coefficients of powers of \( n \), solve for \( A \), \( B \), \( C \), and \( D \). After solving, the values are: \( A = 1 \), \( B = -1 \), \( C = -1 \), \( D = 1 \).
04

Write the Series as a Telescoping Series

Substitute these values back, so:\[ \frac{2n+1}{n^2(n+1)^2} = \frac{1}{n} - \frac{1}{n^2} - \frac{1}{n+1} + \frac{1}{(n+1)^2} \]This form displays the components that cancel between successive terms when summed.
05

Identify the Telescoping Nature

Write out the first few terms of the series using this breakupand you'll see:- Terms \( \frac{1}{n} \) and \( \frac{-1}{n+1} \)- Terms \( \frac{-1}{n^2} \) and \( \frac{1}{(n+1)^2} \)Observe that many terms will cancel when summed.
06

Compute the Sum of the Series

The telescoping nature implies that most internal terms will cancel out, leading to its sum depending only on the initial surviving terms in the series as \( n \to \infty \). Summation proceeds as:\[ S = \lim_{N \to \infty} \left( 1 - \frac{1}{N+1} + \frac{1}{1^2} - \frac{1}{(N+1)^2} \right) \]Therefore, the sum of the series is \( S = 1 \).

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Key Concepts

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

Telescoping Series
Telescoping series are a special type of series where successive terms cancel each other out neatly. This cancellation makes it easier to find the series鈥� sum, especially for large values.
The key idea is that when you write out the first few terms, you will see a pattern of cancellation. For example, in our series, terms like \( \frac{1}{n} \) and \( -\frac{1}{n+1} \) cancel each with its matching counterpart in the next term.
  • This leaves only the initial terms' contribution when you proceed to infinity (\( n \to \infty \)).
  • Such a pattern simplifies the process by reducing large series into the sum of a few simple remaining pieces.
Telescoping series are powerful because they reduce complexity, turning difficult infinite series problems into manageable calculations.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions.
We use it when trying to simplify components of a series, especially when the denominator is a product of polynomial expressions.
  • First, express the complex fraction in terms of simple fractions whose denominators are the original factors.
  • This involves constants that need to be determined by equating both expressions after expansion.
  • For our exercise, we have \( \frac{2n+1}{n^2(n+1)^2} \), and we decomposed it to \( \frac{1}{n} - \frac{1}{n^2} - \frac{1}{n+1} + \frac{1}{(n+1)^2} \).
This simplification not only helps in understanding the nature of each term but also reveals any telescoping patterns hidden in the original series.
Convergence of Series
Convergence refers to the behavior of a series as the number of terms grows to infinity. It determines whether the series approaches a finite value or diverges to infinity.
If a series converges, we can calculate this finite sum, as seen in our example where the sum is 1.
  • The telescoping nature helps ensure that most terms cancel out, leaving only the first few terms to determine the sum.
  • If a series doesn't converge, or the sequence of partial sums doesn't approach a finite value, the series is divergent.
  • In our exercise, by leveraging the telescoping property, identifying surviving terms, and showing the limiting process, we confirmed convergence and found the sum.
Understanding convergence is crucial for working with infinite series, ensuring we derive meaningful and correct results.

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Most popular questions from this chapter

Assume that each sequence converges and find its limit. \(a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}\)

a. Suppose that \(f(x)\) is differentiable for all \(x\) in [0,1] and that \(f(0)=0 .\) Define sequence \(\left\\{a_{n}\right\\}\) by the rule \(a_{n}=n f(1 / n)\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0) .\) Use the result in part (a) to find the limits of the following sequences \(\left\\{a_{n}\right\\}\) \(\mathbf{b} \cdot a_{n}=n \tan ^{-1} \frac{1}{n} \quad\) c. \(a_{n}=n\left(e^{1 / n}-1\right)\) d. \(a_{n}=n \ln \left(1+\frac{2}{n}\right)\)

The approximation \(e^{x}=1+x+\left(x^{2} / 2\right)\) is used when \(x\) is small. Use the Remainder Estimation Theorem to estimate the error when \(|x|<0.1\)

By multiplying the Taylor series for \(e^{x}\) and \(\sin x,\) find the terms through \(x^{5}\) of the Taylor series for \(e^{x} \sin x\). This series is the imaginary part of the series for $$ e^{x} \cdot e^{i x}=e^{(1+i) x} $$ Use this fact to check your answer. For what values of \(x\) should the series for \(e^{x} \sin x\) converge?

Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$. a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.
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