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Chapter 8: Problem 78

It is known that \(\sum_{n=1}^{\infty} \frac{n}{2^{n}}=2\) and \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}=\ln (2)\). Supposing that \(a, b,\) and \(c\) are constants, evaluate \(\sum_{n=1}^{\infty} \frac{a n^{2}+c n+b}{n 2^{n}}\)

Short Answer

Expert verified
The series evaluates to \(2a + c + b \ln(2)\).

Step by step solution

01

Expand the Expression

We are given the series \(\sum_{n=1}^{\infty} \frac{a n^{2}+c n+b}{n 2^{n}}\). We can split the expression into three separate sums: \(a \sum_{n=1}^{\infty} \frac{n^{2}}{n 2^{n}} + c \sum_{n=1}^{\infty} \frac{n}{n 2^{n}} + b \sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\). Simplifying, this becomes \(a \sum_{n=1}^{\infty} \frac{n}{2^{n}} + c \sum_{n=1}^{\infty} \frac{1}{2^{n}} + b \sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\).
02

Substitute Known Values

We are given that \(\sum_{n=1}^{\infty} \frac{n}{2^{n}} = 2\) and \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}} = \ln (2)\). Additionally, the series \(\sum_{n=1}^{\infty} \frac{1}{2^{n}}\) is a geometric series with a sum of 1. Substitute these values into the expression to get: \(a \cdot 2 + c \cdot 1 + b \cdot \ln(2)\).
03

Simplify the Expression

The expression becomes \(2a + c + b \ln(2)\). This is the sum of the series \(\sum_{n=1}^{\infty} \frac{a n^{2}+c n+b}{n 2^{n}}\).

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Key Concepts

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

Geometric Series
A geometric series is a series of numbers with a constant ratio between successive terms. The general form of a geometric series is:
  • First term: \( a \)
  • Common ratio: \( r \)
The sum of an infinite geometric series, where \( |r| < 1 \), is given by the formula \[ S = \frac{a}{1-r} \]. This is because the terms get progressively smaller, approaching zero, as the series progresses.
In this exercise, the series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) is an example of a geometric series. Here, the first term \( a \) is 1 (i.e., \( \frac{1}{2^1} \)), and the common ratio \( r \) is \( \frac{1}{2} \). By applying the formula, the sum converges to 1, affirming the nature of geometric series.
Convergence of Series
Understanding the convergence of series is crucial when dealing with infinite series. A series is said to converge if its sequence of partial sums approaches a specific value, known as the limit. The terms of the series must become increasingly smaller and tend towards zero for convergence to occur.
For instance, the series \( \sum_{n=1}^{\infty} \frac{n}{2^n} = 2 \) converges because, as \( n \) increases, the division by \( 2^n \) overpowers the linear growth of \( n \). Hence, each term in the series becomes smaller, leading to the sum converging to a precise number. Similarly, the series \( \sum_{n=1}^{\infty} \frac{1}{n2^n} = \ln(2) \) also converges due to the rapid growth of the denominator.
  • If the terms of a series do not tend towards zero, the series cannot converge.
  • Geometric series and some power series serve as classic examples of convergence.
Sum of Series
The sum of a series, particularly infinite ones, entails an important concept in calculus and mathematical analysis. It involves determining the total value that the series accumulates to.
In the original exercise, various sums are known or calculated:
  • \( \sum_{n=1}^{\infty} \frac{n}{2^n} = 2 \)
  • \( \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 \)
  • \( \sum_{n=1}^{\infty} \frac{1}{n2^n} = \ln(2) \)
These series are evaluated by combining known sums and manipulating expressions to simplify the computation.
In practice, understanding and applying these principles can simplify finding the sum of complex series, as shown in the exercise, where combining these known sums yields the result \( 2a + c + b\ln(2) \). This is how mathematics beautifully merges theory with precise calculations.

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Most popular questions from this chapter

In Exercises 95-98, use a Taylor polynomial to calculate the given integral to five decimal places. $$ \int_{0}^{1 / 2} e^{-x 2} d x $$

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. Suppose that the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges on \((-R, R)\) to a function \(f(x)\) and that \(|f(x)| \geq k>0\) on that interval for some positive constant \(k\). Then, \(1 / f(x)\) also has a convergent power series expansion on \((-R, R) .\) Compute its coefficients in terms of the \(a_{n}\) 's. Hint: Set $$ \frac{1}{f(x)}=g(x)=\sum_{n=0}^{\infty} b_{n} x^{n} $$ Use the equation \(f(x) \cdot g(x)=1\) to solve for the \(b_{n}\) 's.

In each of Exercises 55-60, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{\sin (x)-x}{(1-\cos (x)) \cdot \ln (1+x)} $$

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 \(\lim _{N \rightarrow \infty} S_{N}\). $$ \sum_{n=1}^{\infty} \frac{1}{(2 n+1)(2 n+3)} $$

A convergent series is given. Estimate the value \(\ell\) of the series by calculating its partial sums \(S_{N}\) for \(N=1,2,3, \ldots\) Round your evaluations to four decimal places and stop when three consecutive rounded partial sums agree. (This procedure does not ensure that the last partial sum calculated agrees with \(\ell\) to four decimal places. The error that results when a partial sum is used to approximate an infinite series is called a truncation error. Methods of estimating truncation errors will be discussed in later sections.) $$ \sum_{n=1}^{\infty} \frac{2^{n}+n^{2}}{10^{n}} $$
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