Problem 3 - Active Calculus

Question

Determine the sum of the following series. $$ \sum_{n=1}^{\infty}\left(\frac{3^{n}+5^{n}}{9^{n}}\right) $$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{7}{4} \).

Step 1: Simplify the General Term of the Series

Consider the general term of the series: \ \( \frac{3^n + 5^n}{9^n} \) \ Notice that \( 9^n = (3^2)^n = (3^n)^2 \), so we can rewrite the term as: \ \( \frac{3^n + 5^n}{(3^n)^2} = \frac{3^n}{(3^n)^2} + \frac{5^n}{(3^n)^2} = \frac{3^n}{3^{2n}} + \frac{5^n}{3^{2n}} \) \ Simplify the exponents in each fraction: \ \( \frac{3^n}{3^{2n}} = 3^{n-2n} = 3^{-n} \) and \( \frac{5^n}{3^{2n}} = \left( \frac{5}{3^2} \right)^n = \left( \frac{5}{9} \right)^n \) \ Thus, the general term becomes: \ \( 3^{-n} + \left( \frac{5}{9} \right)^n \)

Step 2: Separate the Series

The series can now be written as the sum of two separate series: \ \( \sum_{n=1}^{\infty} \left( 3^{-n} + \left( \frac{5}{9} \right)^n \right) = \sum_{n=1}^{\infty} 3^{-n} + \sum_{n=1}^{\infty} \left( \frac{5}{9} \right)^n \)

Step 3: Identify the Geometric Series

Recognize that both parts are geometric series. The first series is: \ \( \sum_{n=1}^{\infty} 3^{-n} \) \ And the second series is: \ \( \sum_{n=1}^{\infty} \left( \frac{5}{9} \right)^n \)

Step 4: Apply the Geometric Series Sum Formula

For a geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum is \( \frac{a}{1-r} \) if \( |r| < 1 \). Here, both series start from \( n=1 \), so note that the general formula uses \( n = 0 \). \ For the first series with \( a = 1/3 \) and \( r = 1/3 \): \ \( \sum_{n=0}^{\infty} 3^{-n} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2} \) \ BUT, we are interested from \( n=1 \): \ Correcting for starting at \( n = 1 \): \ \( \sum_{n=1}^{\infty} 3^{-n} = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2} \) \ For the second series with \( a = 5/9 \) and \( r = 5/9 \): \ \( \sum_{n=0}^{\infty} \left( \frac{5}{9} \right)^n = \frac{1}{1 - 5/9} = \frac{9}{4} \) \ BUT, starting from \( n = 1 \): \ \( \sum_{n=1}^{\infty} \left( \frac{5}{9} \right)^n = \frac{5}{9} \cdot \frac{9}{4} = \frac{5}{4} \)

Step 5: Find the Total Sum

Now, add the sums of the two series together: \ \( \frac{1}{2} + \frac{5}{4} = \frac{2}{4} + \frac{5}{4} = \frac{7}{4} \) \ Therefore, the sum of the series is \( \frac{7}{4} \).

Key Concepts

geometric series

A geometric series is a series of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio, often denoted as 'r'. These series are particularly interesting in mathematics because they have a simple formula for calculating their sum.
For example, the series: 1, 2, 4, 8, ... is a geometric series with a common ratio of 2 since each term is twice the previous term.
The general form of a geometric series can be written as:
\(\text{a} + \text{ar} + \text{ar}^2 + \text{ar}^3 + ... \) where 'a' is the first term and 'r' is the common ratio.
Understanding geometric series is crucial, especially when dealing with more complex series like infinite series.

infinite series

An infinite series is a sum of an infinite sequence of terms. Understanding infinite series is vital because many mathematical concepts and real-world phenomena can be expressed as such series.
The idea is to sum an endless list of terms according to a specific rule, but not just any series will have a finite sum. For instance, the geometric series we looked at earlier can extend infinitely: \(\text{a} + \text{ar} + \text{ar}^2 + ...\).
A crucial condition for the sum of an infinite geometric series to exist is that the common ratio 'r' must satisfy \(|r| < 1\). This condition ensures that the terms of the series get smaller and smaller, approaching zero, which makes it possible for the series to converge to a finite sum.

geometric series sum formula

The geometric series sum formula is a powerful tool that allows us to find the sum of a geometric series. For a geometric series with first term 'a' and common ratio 'r', the formula for the sum 'S' of an infinite geometric series (starting from n=0) is given by:
\[ S = \frac{a}{1-r} \] provided \(|r| < 1\).
In the case of our problem, we used this formula to find the sums separately for \(3^{-n}\) and \(\left( \frac{5}{9} \right)^n\).
For example, for the series \( \sum_{n=0}^{otinfty} 3^{-n}\), with 'a' as 1 and 'r' as 1/3, the sum is:
\[ \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2} \] To correct for starting at n=1, we multiply by 'r' once more to get: \[ \sum_{n=1}^{notinfty} 3^{-n} = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2} \] Similar steps apply for other series combinations. This formula is concise but crucial for understanding and finding the sum of geometric series.