Problem 82 - Single Variable Calculus

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. \(\sum_{n=p}^{\infty} a r^{n}=\frac{a r^{p}}{1-r}\) provided that \(|r|<1\).

Step-by-Step Solution

Answer

The statement is true. The sum of an infinite geometric series, starting at index \(n=p\), is indeed given by \(\sum_{n=p}^{\infty} ar^n = \frac{a r^p}{1-r}\), provided that \(|r|<1\).

Step 1: Recall the infinite geometric series formula.

Consider an infinite geometric series with first term \(a\) and common ratio \(r\). The sum of this series is given by the formula: \[\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, \text{ provided that } |r|<1\]

Step 2: Compare the given statement with the known formula.

The given statement is: \[\sum_{n=p}^{\infty} ar^n = \frac{a r^p}{1-r}, \text{ provided that } |r|<1\] Notice that this sum starts from index \(n=p\) instead of \(n=0\). We can rewrite the sum as follows: \[\sum_{n=p}^{\infty} ar^n = a r^p \sum_{n=0}^{\infty} ar^n\]

Step 3: Check if the sum is equal to the given statement.

Now, we can plug in the formula for the sum of the infinite geometric series: \[\sum_{n=p}^{\infty} ar^n = a r^p \sum_{n=0}^{\infty} ar^n = a r^p \frac{a}{1-r}\] This equation matches the given statement, \(\sum_{n=p}^{\infty} ar^n = \frac{a r^p}{1-r}\).

Step 4: Conclusion

The statement is true. The sum of an infinite geometric series, starting at index \(n=p\), is indeed given by \(\sum_{n=p}^{\infty} ar^n = \frac{a r^p}{1-r}\), provided that \(|r|<1\).

Key Concepts

Convergence of Series

Understanding the convergence of series is essential for handling problems involving infinite series. Convergence refers to the behavior of an infinite series as the number of terms increases indefinitely. A series is said to converge if the sum of its terms approaches a specific finite value as the number of terms grows. Conversely, if the sum continues to grow without bound or fails to settle down to a number, the series diverges.

For example, the statement \(\sum_{n=p}^{\infty} a r^n = \frac{a r^p}{1-r}\) implies that the geometric series converges, given the condition that \(|r|<1\). This condition is critical; the absolute value of the common ratio \(r\) must be less than one for the series to converge. If \(r\) is greater or equal to one in magnitude, the series terms grow larger with each addition, leading to divergence. When dealing with an infinite geometric series, always check the modulus of the common ratio to determine convergence.

Geometric Series Sum Formula

The sum formula for an infinite geometric series is a powerful tool when solving problems related to series convergence. For a series with first term \(a\) and common ratio \(r\), the geometric series sum formula is \[\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r},\] providing that \(|r|<1\). This formula derives from the concept that each term in the series is the previous term multiplied by the constant \(r\).

If we are to consider a series that begins with an index other than zero, such as \(p\), we adjust the formula by multiplying the standard formula by \(r^p\). This results in the correctly adjusted sum of \(\frac{a r^p}{1-r}\). It's crucial to understand this formula and its derivation, as it is the basis for calculating the sum of any infinite geometric series where the common ratio falls strictly between -1 and 1.

Series Index Shift

Index shifting in series is a concept that allows the adaptation of a general formula to an altered starting point. Typically, geometric series are expressed with a starting index of \(n=0\). However, an index shift may occur, and the series might begin at any arbitrary \(n=p\). When this happens, we need to adjust our approach to compute the sum correctly.

Series index shift does not change the series' nature but simply shifts the 'starting line.' To account for this, one can factor out the terms that make up the shift. For instance, a shift involving the index \(p\) can be addressed by noting that:

• \[\sum_{n=p}^{\infty} ar^n = ar^p \sum_{n=0}^{\infty} r^n\]

This transformation uses the properties of exponents to factor out \(ar^p\) as the first \(p\) terms of the series. When dealing with shifted series, always redistribute the sum to reflect the starting index accurately before applying the geometric series sum formula.