Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(1-14\), indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series $$ \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} $$

Step-by-Step Solution

Answer

The series converges with a sum of \( \frac{1}{6} \).

Step 1: Identify the Series Type

The given series \( \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} \) is a geometric series. A geometric series has the general form \( \sum_{k=0}^{\infty} ar^{k} \) or \( \sum_{k=1}^{\infty} ar^{k-1} \), where \( a \) is the first term and \( r \) is the common ratio.

Step 2: Determine the First Term and Common Ratio

For the series \( \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} \), identify \( a \) and \( r \). Since the series starts at \( k=1 \), the first term \( a = \left(\frac{1}{7}\right)^{1} = \frac{1}{7} \), and the common ratio \( r = \frac{1}{7} \).

Step 3: Check Convergence of the Series

A geometric series converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( r = \frac{1}{7} \) and \( |\frac{1}{7}| < 1 \), so the series converges.

Step 4: Calculate the Sum of the Series

Since the series converges, use the formula for the sum of an infinite geometric series: \( S = \frac{a}{1-r} \). Substituting the values: \( a = \frac{1}{7} \) and \( r = \frac{1}{7} \), the sum \( S = \frac{\frac{1}{7}}{1-\frac{1}{7}} = \frac{\frac{1}{7}}{\frac{6}{7}} = \frac{1}{6} \).

Step 5: Conclusion

The series \( \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} \) converges, and its sum is \( \frac{1}{6} \).

Key Concepts

Convergence

Convergence is a key concept when dealing with infinite series, especially geometric ones. To understand convergence, think of it as determining whether a series can "settle down" to a specific value or not. Specifically, a series converges if the sum of its terms approaches a finite number as the number of terms grows very large.
A geometric series will converge if its common ratio, denoted by \( r \), satisfies \( |r| < 1 \). In such cases, the larger the series gets, the closer the sum reaches a particular finite value.
In our example series \( \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} \), the common ratio \( r = \frac{1}{7} \), which satisfies \( |r| < 1 \). That's why the series converges. If the common ratio had been greater than or equal to 1 in absolute value, the series would not have converged and the sum would have continued indefinitely.

Infinite Series

An infinite series is the sum of an infinite sequence of numbers. The idea is to keep adding terms one after another, seemingly never-ending. In theory, infinite series are fascinating because they explore sums that go on without stopping.
However, not all infinite series make sense or come to a meaningful conclusion. Some series, called divergent series, just keep growing larger and larger without settling on a specific value.

• A geometric series, which we discussed, is a specific type of infinite series where each term after the first is found by multiplying the previous term by a constant, the common ratio \( r \).
• For example, in \( \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} \), every term is obtained by multiplying the previous term by \( \frac{1}{7} \).

Understanding whether an infinite series converges or diverges is crucial, as it helps us know whether an infinite sum can actually reach a certain point.

Sum of a Series

Once we've established that an infinite series converges, the next step is to calculate its sum. This tells us the exact value that the series approaches as it grows infinitely large.
For a geometric series, if the absolute value of the common ratio \( |r| < 1 \), we can use a simple formula to find the sum:
\[ S = \frac{a}{1 - r} \]

• \( a \) is the first term of the series.
• \( r \) is the common ratio.

In the series \( \sum_{k=1}^{\infty}\left(\frac{1}{7}\right)^{k} \), we found that \( a = \frac{1}{7} \) and \( r = \frac{1}{7} \). Plugging these into the formula gives us:
\[ S = \frac{\frac{1}{7}}{1 - \frac{1}{7}} = \frac{\frac{1}{7}}{\frac{6}{7}} = \frac{1}{6} \]
This calculation shows that no matter how many terms we add, the sum will always approach \( \frac{1}{6} \). This result makes infinite series both puzzling and beautiful, revealing hidden patterns in seemingly never-ending sequences.