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Chapter 9: Problem 82

Find all values of \(x\) for which the series converges. For these values of \(x\), write the sum of the series as a function of \(x\). $$ \sum_{n=0}^{\infty} 4\left(\frac{x-3}{4}\right)^{n} $$

Short Answer

Expert verified
For the given series, it is convergent and its sum can be expressed as a function of \(x\) when \(x\) is between 1 and 7 (exclusive). In these cases, the sum is given by \(S = 16 / (7 - x)\).

Step by step solution

01

Identify the Series Type and Parameters

The given series is a geometric series. The common ratio \(a\) can be identified as \((x-3)/4\) and the first term \(a_0\) of the series is when \(n=0\), that is, \(4\).
02

Find the Convergence Condition

For a geometric series, the series converges if the absolute value of the ratio is less than one. This gives the condition as, \(| (x - 3)/4 | < 1\). From here, we can unravel the inequality to determine the range of \(x\).
03

Solve for \(x\)

Solving the inequality for \(x\), we get \(x\) is greater than 1 and less than 7.
04

Write the Sum as a Function of \(x\)

For the range of \(x\) where the series converges, the sum of the series can be written using the formula of sum of an infinite geometric series, \(S = a_0 / (1 - a)\). Substituting the values of \(a_0\) and \(a\), we get the sum \(S = 16 / (7 - x)\).

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Key Concepts

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

Convergence
Convergence occurs when the terms of a series get closer and closer to a specific value as the number of terms increases. In a geometric series, we find convergence by analyzing the common ratio. The series converges if and only if the absolute value of this ratio is less than one. Therefore, for the series \[\sum_{n=0}^{\infty} 4\left(\frac{x-3}{4}\right)^{n}\]to converge, we need \[\left| \frac{x-3}{4} \right| < 1.\]This inequality helps us determine the range of values for which the series converges by solving for the variable \(x\). Ultimately, convergence ensures that we can calculate the sum of the series.
Sum of Series
The sum of a convergent geometric series can be found using a specific formula. For the given series, once we establish that it converges, calculating the sum becomes straightforward. When a geometric series converges, its sum \(S\) is given by \[S = \frac{a_0}{1 - a}\]where \(a_0\) is the first term, and \(a\) is the common ratio. In our exercise, \(a_0 = 4\) and \(a = \frac{x-3}{4}\). Thus, the sum of the series is \[S = \frac{4}{1 - \frac{x-3}{4}} = \frac{16}{7-x}\]for any \(x\) within the convergence range. Knowing this formula allows us to directly calculate the sum for different values of \(x\).
Inequality
Inequality plays a crucial role in determining the convergence of our geometric series. It involves establishing a range for \(x\) by solving \[\left| \frac{x-3}{4} \right| < 1.\]To solve this inequality, break it into two parts:
  • \(\frac{x-3}{4} < 1\)
  • \(\frac{x-3}{4} > -1\)
Solving these, we find that:
  • \(x-3 < 4\) leads to \(x < 7\)
  • \(x-3 > -4\) leads to \(x > 1\)
Thus, combining these inequalities gives us the range \(1 < x < 7\), which is critical for the series to converge. Understanding this concept helps verify when the sum formula is applicable.
Geometric Series Formula
The geometric series formula is a powerful tool in calculus to calculate the sum of infinite series. For a geometric series, if the series converges, the sum can be determined by the formula \[S = \frac{a_0}{1 - a}\]where \(a_0\) is the initial term of the series and \(a\) represents the common ratio. This formula highlights how quickly we can ascertain the full sum even for an infinite series, as long as it meets the convergence criteria. Applied to our problem, it simplifies finding the sum for any valid \(x\). Recognizing a geometric series and correctly using this formula enables quick decision-making in mathematics.

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Most popular questions from this chapter

Consider the sequence \(\left\\{a_{n}\right\\}\) where \(a_{1}=\sqrt{k}, a_{n+1}=\sqrt{k+a_{n}}\), and \(k>0\). (a) Show that \(\left\\{a_{n}\right\\}\) is increasing and bounded. (b) Prove that \(\lim _{n \rightarrow \infty} a_{n}\) exists. (c) Find \(\lim _{n \rightarrow \infty} a_{n^{\prime}}\)

The series represents a well-known function. Use a computer algebra system to graph the partial sum \(S_{10}\) and identify the function from the graph. $$ f(x)=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !} $$

Find all values of \(x\) for which the series converges. For these values of \(x\), write the sum of the series as a function of \(x\). $$ \sum_{n=1}^{\infty}(x-1)^{n} $$

Find the Maclaurin series for \(f(x)=\ln \frac{1+x}{1-x}\) and determine its radius of convergence. Use the first four terms of the series to approximate \(\ln 3 .\)

Use the formula for the \(n\) th partial sum of a geometric series \(\sum_{i=0}^{n-1} a r^{i}=\frac{a\left(1-r^{n}\right)}{1-r}\) Annuities When an employee receives a paycheck at the end of each month, \(P\) dollars is invested in a retirement account. These deposits are made each month for \(t\) years and the account earns interest at the annual percentage rate \(r\). If the interest is compounded monthly, the amount \(A\) in the account at the end of \(t\) years is $$ \begin{aligned} A &=P+P\left(1+\frac{r}{12}\right)+\cdots+P\left(1+\frac{r}{12}\right)^{12 t-1} \\ &=P\left(\frac{12}{r}\right)\left[\left(1+\frac{r}{12}\right)^{12 t}-1\right] \end{aligned} $$ If the interest is compounded continuously, the amount \(A\) in the account after \(t\) years is $$ \begin{aligned} A &=P+P e^{r / 12}+P e^{2 r / 12}+P e^{(12 t-1) r / 12} \\ &=\frac{P\left(e^{n}-1\right)}{e^{r / 12}-1} \end{aligned} $$ Verify the formulas for the sums given above.
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