Problem 12 - Thomas Calculus

Question

In Exercises \(7-14,\) write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. $$ \sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}-\frac{1}{3^{n}}\right) $$

Step-by-Step Solution

Answer

The series converges with a sum of \(\frac{23}{2}\).

Step 1: Identify the Series Components

The given series is \[ \sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}-\frac{1}{3^{n}}\right). \]There are two separate geometric series within this series:1. \( \left(\frac{5}{2^{n}}\right) \)2. \( \left(-\frac{1}{3^{n}}\right) \)

Step 2: Write Out the First Eight Terms of Each Component

Calculate the first eight terms for each component separately:For \( \frac{5}{2^{n}} \): \\[ n=0: \frac{5}{1}=5, \ n=1: \frac{5}{2}, \ n=2: \frac{5}{4}, \ n=3: \frac{5}{8}, \ n=4: \frac{5}{16}, \ n=5: \frac{5}{32}, \ n=6: \frac{5}{64}, \ n=7: \frac{5}{128}. \]For \( \left(-\frac{1}{3^{n}}\right) \):\[ n=0: \frac{-1}{1}=-1, \ n=1: \frac{-1}{3}, \ n=2: \frac{-1}{9}, \ n=3: \frac{-1}{27}, \ n=4: \frac{-1}{81}, \ n=5: \frac{-1}{243}, \ n=6: \frac{-1}{729}, \ n=7: \frac{-1}{2187}. \]

Step 3: Compute the First Eight Terms of the Combined Series

Add the corresponding terms from the two series to get the terms of the combined series:- 1st term: \( 5 - 1 = 4 \)- 2nd term: \( \frac{5}{2} - \frac{1}{3} = \frac{15}{6} - \frac{2}{6} = \frac{13}{6} \)- 3rd term: \( \frac{5}{4} - \frac{1}{9} = \frac{45}{36} - \frac{4}{36} = \frac{41}{36} \)- 4th term: \( \frac{5}{8} - \frac{1}{27} = \frac{135}{216} - \frac{8}{216} = \frac{127}{216} \)- 5th term: \( \frac{5}{16} - \frac{1}{81} = \frac{405}{1296} - \frac{16}{1296} = \frac{389}{1296} \)- 6th term: \( \frac{5}{32} - \frac{1}{243} = \frac{1215}{7776} - \frac{32}{7776} = \frac{1183}{7776} \)- 7th term: \( \frac{5}{64} - \frac{1}{729} = \frac{3645}{46656} - \frac{64}{46656} = \frac{3581}{46656} \)- 8th term: \( \frac{5}{128} - \frac{1}{2187} = \frac{10935}{279936} - \frac{128}{279936} = \frac{10807}{279936} \)

Step 4: Analyze the Convergence of Each Component

Each component is an infinite sum of a geometric series. \- The series \( \sum_{n=0}^{\infty} \frac{5}{2^n} \) has a common ratio \( r = \frac{1}{2} \). Since \( |r| < 1 \), it converges, and its sum is \( \frac{5}{1 - \frac{1}{2}} = 10 \).- The series \( \sum_{n=0}^{\infty} -\frac{1}{3^n} \) also has a common ratio \( r = \frac{1}{3} \), so it converges with sum \( \frac{-1}{1 - \frac{1}{3}} = -\frac{3}{2} \).

Step 5: Calculate the Sum of the Series

Since both series converge, we can find the sum of the original series:\[ \sum_{n=0}^{\infty} \left( \frac{5}{2^n} - \frac{1}{3^n} \right) = \sum_{n=0}^{\infty} \frac{5}{2^n} - \sum_{n=0}^{\infty} \frac{1}{3^n}. \]Thus, the total sum is \[ 10 - \left(-\frac{3}{2}\right) = 10 + \frac{3}{2} = \frac{23}{2}. \]

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. It's an essential concept in mathematics that helps us understand and analyze patterns.
Consider the series \( \sum_{n=0}^{\infty} ar^n \), where \(a\) is the first term and \(r\) the common ratio. A geometric series can be finite or infinite. If the absolute value of the common ratio is less than 1 (\(|r| < 1\)), the series will converge, meaning it approaches a specific sum as the number of terms increases.

• For example, for the series \( \sum_{n=0}^{\infty} \frac{5}{2^n} \), the common ratio \( r = \frac{1}{2} \).
• For the series \( \sum_{n=0}^{\infty} -\frac{1}{3^n} \), the common ratio \( r = \frac{1}{3} \).

Both series converge because \(|r| < 1\) in each case. Understanding geometric series allows us to efficiently find their summation, provided they converge.

Sum of Infinite Series

The sum of an infinite series, when it converges, is found using the formula: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term, and \( r \) is the common ratio of the series.
When working with geometric series like \( \sum_{n=0}^{\infty} ar^n \), this formula is invaluable for quickly reaching the sum when \(|r| < 1\).
For our initial series, we have two separate sums to calculate:

• The sum of \( \sum_{n=0}^{\infty} \frac{5}{2^n} \) is \( \frac{5}{1 - \frac{1}{2}} = 10 \).
• The sum of \( \sum_{n=0}^{\infty} -\frac{1}{3^n} \) is \( \frac{-1}{1 - \frac{1}{3}} = -\frac{3}{2} \).

The total sum of these two series is simply the sum of their two individual sums:
\( 10 - \left(-\frac{3}{2}\right) = 10 + \frac{3}{2} = \frac{23}{2} \).
Thus, the sum of the original infinite series can be quickly calculated once we verify that each sub-series converges.

Sequence and Series Analysis

Sequence and series analysis is all about understanding how sequences and their sums behave. Sequences are ordered lists of numbers, and when we talk about series, we're referring to sums of these sequences.
In the case of geometric series, analysis is particularly focused on convergence, which tells us if the sum of all terms leads to a finite number. To identify convergence, check if the absolute value of the common ratio is less than 1 (\(|r| < 1\)).

• Our exercise involves analyzing the series \( \sum_{n=0}^{\infty}(\frac{5}{2^n} - \frac{1}{3^n}) \).
• This involves identifying two sub-series, checking their convergence, and combining their sums.

Performing sequence and series analysis helps in simplifying complex problems by breaking them down into manageable parts. You start with identifying smaller components or sub-series, analyze them individually, calculate their sums, and then determine the overall behavior of the whole series. This holistic understanding makes these mathematical concepts much clearer.