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Chapter 9: Problem 29

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{4^{n}}{3^{n}+1} $$

Short Answer

Expert verified
According to the Ratio Test, the series \( \sum_{n=0}^{\infty} \frac{4^{n}}{3^{n}+1} \) converges.

Step by step solution

01

Note the given series

The given series is \[\sum_{n=0}^{\infty} \frac{4^{n}}{3^{n}+1}\]
02

Formulate the ratio of the (n+1)th term to the nth term

The ratio of the (n+1)th term to the nth term would be \[\left|\frac{(4^{n+1}/(3^{n+1}+1))}{(4^{n}/(3^{n}+1))}\right|\]
03

Simplify the ratio

Now, simplifying the ratio, we get \[\left|\frac{4 \cdot 3^n \cdot 3}{4 \cdot 3^n}\right| = \left|\frac{3}{4}\right|\]after cancelling terms.
04

Evaluate the limit

Calculate the limit of this ratio as n approaches infinity.\[\lim_{n\to\infty} \left|\frac{3}{4}\right|\]The limit does not depend on n and is therefore \( \frac{3}{4} \).
05

Determine convergency

To determine the convergence of the series, we apply the Ratio Test. As our limit is \( \frac{3}{4} \), which is less than 1, the series converges according to the Ratio Test.

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Key Concepts

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

Convergence of Series
When analyzing an infinite series using the Ratio Test, one main goal is to determine whether the series converges. Convergence implies that as you add more and more terms to the series, the sum approaches a finite number. This is crucial since only convergent series can represent meaningful sums or functions in many mathematical contexts, like convergent power series used in calculus for functions approximations.
To apply the Ratio Test, we find the ratio of the \(n+1\)th term to the nth term of the series and then take its limit as \(n\) approaches infinity. If this limit is less than 1, the series converges.
For example, in the given series, \\[\sum_{n=0}^{\infty} \frac{4^{n}}{3^{n}+1}\]\ The ratio obtained after simplification is \(\frac{3}{4}\), which is less than 1. Hence, the series converges according to the Ratio Test.
Divergence of Series
When discussing series, divergence refers to a scenario where the series doesn't settle at a certain value and instead, either grows indefinitely or oscillates without settling. For a series to be meaningful in many mathematical applications, it usually needs to converge.
The Ratio Test is a handy tool to explore divergence. After finding the limit of the ratio of successive terms, if the limit is greater than 1, the series diverges. In such cases, the sum doesn't add up to a particular value, making it less applicable in real-world calculations.
Using this idea, the given series \[\sum_{n=0}^{\infty} \frac{4^{n}}{3^{n}+1}\] provides a limit ratio of \(\frac{3}{4}\) after simplification, which is less than 1, indicating convergence and not divergence. Had the ratio exceeded 1, then the series would have diverged.
Infinite Series
Infinite series involve adding up infinitely many terms. They're expressed in the form \[\sum_{n=0}^{\infty} a_n\]\ Here, \(a_n\) is the general term of the series. An infinite series is a central concept in mathematics because they can represent decimal expansions, functions, and solutions of differential equations.
Understanding whether an infinite series converges or diverges is crucial. Convergent infinite series approximate numbers or functions with unlimited accuracy. For the given series, \[\sum_{n=0}^{\infty} \frac{4^{n}}{3^{n}+1}\], infinity indicates endless addition of terms. Yet, since it converges, these sums taper off to a finite limit.
  • Infinite series are foundational in calculus.
  • They aid in studying sequences and summations.
  • The Ratio Test is valuable for determining the behavior of such series.
Recognizing the nature of infinite series and their convergence ties back directly to their applications and the power intrinsic to their use in advanced calculus and analysis.

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Most popular questions from this chapter

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 \(x\). $$ \sum_{n=0}^{\infty}(-1)^{n} x^{n} $$

Show that the Maclaurin series of the function \(g(x)=\frac{x}{1-x-x^{2}}\) is \(\sum_{n=1}^{\infty} F_{n} x^{n}\) where \(F_{n}\) is the \(n\) th Fibonacci number with \(F_{1}=F_{2}=1\) and \(F_{n}=F_{n-2}+F_{n-1}\), for \(n \geq 3 .\) (Hint: Write \(\frac{x}{1-x-x^{2}}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots\) and multiply each side of this equation by \(1-x-x^{2}\).)

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.

Arithmetic-Geometric Mean Iet \(a_{0}>b_{0}>0\). I.et \(a_{1}\) be the arithmetic mean of \(a_{0}\) and \(b_{0}\) and let \(b_{1}\) be the geometric mean of \(a_{0}\) and \(b_{0}\). \(a_{1}=\frac{a_{0}+b_{0}}{2} \quad\) Arithmetic mean \(b_{1}=\sqrt{a_{0} b_{0}} \quad\) Geometric mean Now define the sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) as follows. \(a_{n}=\frac{a_{n-1}+b_{n-1}}{2} \quad b_{n}=\sqrt{a_{n-1} b_{n-1}}\) (a) Let \(a_{0}=10\) and \(b_{0}=3 .\) Write out the first five terms of \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\} .\) Compare the terms of \(\left\\{b_{n}\right\\} .\) Compare \(a_{n}\) and \(b_{n}\). What do you notice? (b) Use induction to show that \(a_{n}>a_{n+1}>b_{n+1}>b_{n}\), for \(a_{0}>b_{0}>0\) (c) Explain why \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) are both convergent. (d) Show that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}\).

show that the function represented by the power series is a solution of the differential equation. $$ y=1+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{4 n}}{2^{2 n} n ! \cdot 3 \cdot 7 \cdot 11 \cdot \cdots(4 n-1)}, y^{\prime \prime}+x^{2} y=0 $$
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