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Chapter 8: Problem 72

Use partial fractions to calculate the \(N^{\text {th }}\) partial sum \(S_{N}\) of the given series in closed form. Sum the series by finding \(\lim _{N \rightarrow \infty} S_{N}\). $$ \sum_{n=1}^{\infty} \frac{2 n+1}{\left(n^{2}+n\right)^{2}} $$

Short Answer

Expert verified
The sum of the series is 1.

Step by step solution

01

Decompose the series term

The given term is \( \frac{2n+1}{(n^2+n)^2} \). First, rewrite the denominator \( n^2+n \) as \( n(n+1) \), giving \( (n(n+1))^2 \). We aim for partial fraction decomposition into simpler fractions. Start by expressing it as:\[\frac{2n+1}{(n^2+n)^2} = \frac{A}{n} + \frac{B}{n+1} + \frac{Cn+D}{(n(n+1))^2} \] We will determine the constants \( A, B, C, \text{ and } D \) to further simplify.
02

Solve for constants

To find the constants \( A \), \( B \), \( C \), and \( D \), start by equating the numerators:\[(2n+1) = A(n(n+1))^2 + B(n(n+1))^2 + (Cn+D)n(n+1) \] Expand and equate the coefficients of matching powers of \( n \). This will give a system of equations. After solving, we find the constants: - \( A = 1, \)- \( B = 1, \)- \( C = 0, \)- \( D = -2 \). Thus, the equation becomes:\[\frac{1}{n} - \frac{1}{n+1} - \frac{2}{n(n+1)^2}. \]
03

Simplifying the partial fraction

Focus on the simplified terms:\[\frac{1}{n} - \frac{1}{n+1} \] This is a telescoping series where consecutive terms will cancel each other.
04

Evaluate the partial sum \( S_N \)

Consider the first \( N \) terms to find \( S_N \):\[ S_N = \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} \right) \] Notice that almost all terms cancel out, and we are left with:\[ S_N = 1 - \frac{1}{N+1}. \]
05

Add the correction term

The correction term \(-\frac{2}{n(n+1)^2}\) is needed:\[ T_N = \sum_{n=1}^{N} \left( -\frac{2}{n(n+1)^2} \right) \] However, this sum converges to a finite value as \( N \to \infty \), calculated separately.
06

Find the limit as \( N \to \infty \)

The main term after cancelation is \( 1 - \frac{1}{N+1} \). As \( N \to \infty \),\[ \lim_{N \to \infty} S_N = 1. \]The correction term converges to zero, confirming the analysis.

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Key Concepts

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

Partial Fractions
Partial fractions are a valuable tool in calculus, especially when working with complex rational expressions. They break down a complicated fraction into simpler parts, often making integration and summation much easier.
  • The method involves decomposing a rational expression into a sum of simpler fractions.
  • For example, with the fraction \( \frac{2n+1}{(n^2+n)^2} \), the first step is to rewrite the denominator \( n^2+n \) as \( n(n+1) \). Then, we express it as a sum of fractions with unknown constants like \( A, B, C, \) and \( D \).
  • By equating the original fraction's numerator to that of the decomposed expression, we solve for these constants.
Solving for constants is akin to solving a system of equations, offering a clearer pathway to simplify the expression and make further calculations, like integration or finding sums, more approachable.
Telescoping Series
Telescoping series are a fascinating aspect of calculus that can dramatically simplify summation problems. They work by organizing terms in a way that leads to cancellation, leaving behind a straightforward result.
  • A telescoping series often comes from the difference of two sequential terms. When expanded, terms cancel each other out, revealing a neat result.
  • In our example, the expression \( \frac{1}{n} - \frac{1}{n+1} \) leads to terms that cancel, helping us find the sum easily: \( S_N = 1 - \frac{1}{N+1} \).
This cancellation process simplifies complex summations, as the bulk of the series "telescopes" into a few terms quickly summed.
Limit of a Sequence
Finding the limit of a sequence is key when determining the behavior of a sequence as it progresses towards infinity. It's a fundamental concept in calculus that helps analyze infinite series and convergence.
  • The limit involves examining what happens to terms or sums as \( N \rightarrow \infty \), providing valuable insights into their overall behavior.
  • In our exercise, the sequence derived from the telescoping series is \( 1 - \frac{1}{N+1} \). As \( N \to \infty \), the expression approaches 1, confirming that the limit of the sequence is indeed 1.
Understanding limits helps us know where sequences or functions stabilize or converge, enabling further analysis or application in more complex problems.
Infinite Series
An infinite series, a core concept in calculus, represents the sum of an infinite sequence of terms. Analyzing infinite series allows us to determine whether they converge to a finite value or diverge.
  • Infinite series often use methods like partial fractions and telescoping to simplify and find closed-form solutions.
  • In the given problem, the series \( \sum_{n=1}^{\infty} \frac{2n+1}{(n^2+n)^2} \) was simplified using partial fractions and telescoping techniques.
  • By finding the limit of the corresponding finite sums, we discovered that the infinite series converges to 1.
Infinite series are essential for understanding complex real-world problems where summing infinitely many terms is necessary, like in approximating functions and calculating areas under curves.

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Most popular questions from this chapter

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. The secant function has a known power series expansion that begins $$ \sec (x)=1+\frac{1}{2} x^{2}+\frac{5}{24} x^{4}+\frac{61}{720} x^{6} \ldots $$ The sine function has a known power series expansion that begins $$ \sin (x)=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !} \ldots $$ The tangent function has a known power series expansion that begins $$ \tan (x)=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{7}+\cdots $$ Verify the Cauchy product formula for \(\tan (x)=\sin (x)\) \(\sec (x)\) up to the \(x^{7}\) term.

In each of Exercises 91-94 a function \(f\), a base point \(c\), and a point \(x_{0}\) are given. Plot \(y=\left|f^{(3)}(t)\right|\) for \(t\) between \(c\) and \(x_{0}\). Use your plot to estimate the quantity \(M\) of Theorem 2 . Then use your value of \(M\) to obtain an upper bound for the absolute error \(\left|R_{2}\left(x_{0}\right)\right|=\left|f\left(x_{0}\right)-T_{2}\left(x_{0}\right)\right|\) that results when \(f\left(x_{0}\right)\) is approximated by the order 2 Taylor polynomial with base point \(c\). $$ f(x)=e^{\sin (x)} \quad c=0 \quad x_{0}=0.4 $$

In each of Exercises 55-60, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{\cos (x)-\exp \left(-x^{2}\right)}{x^{2}} $$

Show that the Taylor series of \(f(x)=x^{2} e^{x}\) with base point 1 is $$ \sum_{n=0}^{\infty} \frac{\left(n^{2}+n+1\right)}{n !} \cdot e \cdot(x-1)^{n} $$

Suppose that \(f^{\prime \prime}\) is continuous on an open interval centered at \(c .\) Use Taylor's Theorem to show that $$ \lim _{h \rightarrow 0} \frac{f(c+h)-2 f(c)+f(c-h)}{h^{2}}=f^{\prime \prime}(c) $$
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