Problem 31
Question
Determine whether the given series converges or diverges. If it converges, find its sum. \(\sum_{n=0}^{\infty} \frac{2^{n}}{5^{n}}\)
Step-by-Step Solution
Verified Answer
The series \(\sum_{n=0}^{\infty} \frac{2^{n}}{5^{n}}\) converges because its common ratio \(r = \frac{2}{5}\) has an absolute value less than 1. The sum of the series is \(S = \frac{1}{1 - \frac{2}{5}} = \frac{5}{3}\).
1Step 1: Identify the first term and common ratio
Given the series, \(\sum_{n=0}^{\infty} \frac{2^{n}}{5^{n}}\), the general term is \(a_n = \frac{2^n}{5^n}\).
- The first term: \(a_0 = \frac{2^0}{5^0} = \frac{1}{1} = 1\)
- The common ratio: \(r = \frac{2}{5}\)
2Step 2: Check the convergence condition
A geometric series converges if and only if the common ratio \(|r| < 1\). In this case, \(r = \frac{2}{5}\), hence \(|r| = |\frac{2}{5}| = \frac{2}{5}\), since the fraction is positive.
Therefore, using the condition \(\frac{2}{5} < 1\), the series converges.
3Step 3: Find the sum of the series
Since we have proved that the series converges, we can use the sum formula for a geometric series: \(S = \frac{a}{1-r}\).
Substitute the first term and common ratio into the formula:
\(S = \frac{1}{1 - \frac{2}{5}}\)
Finding the common denominator in the denominator of the fraction:
\(S = \frac{1}{\frac{5 - 2}{5}}\)
Simplify the fraction:
\(S = \frac{1}{\frac{3}{5}}\)
To solve this, we can multiply the numerator and denominator by \(5\):
\(S = \frac{1 \cdot 5}{3 \cdot 1}\)
\(S = \frac{5}{3}\)
So, the sum of the series \(\sum_{n=0}^{\infty} \frac{2^{n}}{5^{n}}\) is \(\frac{5}{3}\).
Key Concepts
Series ConvergenceSum of SeriesCommon Ratio Calculation
Series Convergence
In mathematics, series convergence refers to a condition where an infinite series approaches a specific value as more terms are added. Understanding when a series converges is crucial since it allows us to determine if there is a fixed sum we can assign to it.
To identify convergence in geometric series, we focus on the common ratio, usually denoted as \( r \). A geometric series is expressed in the form \( a + ar + ar^2 + ar^3 + \ldots \). For the series to converge, the absolute value of the common ratio, \(|r|\), must be less than 1, i.e., \(|r| < 1\).
If this condition is met, each subsequent term becomes smaller, and the series approaches a sum. If not, the terms either increase or do not diminish sufficiently, moving the series further from a specific value.
To identify convergence in geometric series, we focus on the common ratio, usually denoted as \( r \). A geometric series is expressed in the form \( a + ar + ar^2 + ar^3 + \ldots \). For the series to converge, the absolute value of the common ratio, \(|r|\), must be less than 1, i.e., \(|r| < 1\).
If this condition is met, each subsequent term becomes smaller, and the series approaches a sum. If not, the terms either increase or do not diminish sufficiently, moving the series further from a specific value.
- The series \( \sum_{n=0}^{\infty} \frac{2^n}{5^n} \) has \( r = \frac{2}{5} \).
- Here, \(|r| = \frac{2}{5} = 0.4\) which is less than 1. Therefore, the series converges.
Sum of Series
Once we establish that a geometric series converges, calculating its sum becomes straightforward using a specific formula. For a convergent geometric series, the sum can be found with \( S = \frac{a}{1-r} \), where \( a \) is the first term, and \( r \) is the common ratio.
For our series, the first step was to find the first term (also called the initial term or starting point):
So, the sum of this series is \( \frac{5}{3} \), showing how even infinite sequences can sum to finite numbers when they converge.
For our series, the first step was to find the first term (also called the initial term or starting point):
- Here, \( a = 1 \) since \( a_0 = \frac{2^0}{5^0} = 1 \).
- First, simplify the denominator: \( 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5} \).
- Then adjust the fraction: \( S = \frac{1}{\frac{3}{5}} = \frac{5}{3} \).
So, the sum of this series is \( \frac{5}{3} \), showing how even infinite sequences can sum to finite numbers when they converge.
Common Ratio Calculation
Understanding and calculating the common ratio is a fundamental aspect of analyzing geometric series. The common ratio, \( r \), is essentially the factor by which you multiply one term to get the next. For a series represented by \( a_n = \frac{a_{n+1}}{a_n} \), \( r \) is the constant that connects consecutive terms.
Let's see how we find it for the series \( \sum_{n=0}^{\infty} \frac{2^n}{5^n} \):
The sign and value of \( r \) have both vital roles:
Let's see how we find it for the series \( \sum_{n=0}^{\infty} \frac{2^n}{5^n} \):
- The general term here is \( a_n = \frac{2^n}{5^n} \).
- Therefore, the common ratio is \( r = \frac{2}{5} \), calculated by dividing the numerator and denominator separately to find the repeating factor.
The sign and value of \( r \) have both vital roles:
- If \(|r| < 1\), the series converges; here, \( |\frac{2}{5}| = 0.4 \).
- Specific scenarios might involve a negative \( r \), causing terms to alternate in sign.
Other exercises in this chapter
Problem 31
Use the binomial series to find the power series representation of the function. Then find the radius of convergence of the series. \(f(x)=\sqrt{1-x^{2}}\)
View solution Problem 31
Determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{\sin ^{2} n}{n \sqrt{n+1}}\)
View solution Problem 31
Find an approximation of the sum of the series accurate to two decimal places. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}} $$
View solution Problem 31
Determine whether the given series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}+2 n+5} $$
View solution