Problem 92 - Essential Calculus

Question

Write an equivalent series with the index of summation beginning at \(n=0\). $$ \sum_{n=2}^{\infty} \frac{2^{n}}{(n-2) !} $$

Step-by-Step Solution

Answer

The equivalent series with the index of summation beginning at \(n=0\) is \(\sum_{i=0}^{\infty} \frac{4 \cdot 2^{i}}{i !}\).

Step 1: Translate the Index

First, introduce a new index \(i\) where \(i=n-2\). Therefore, \(n=i+2\). Using this new index, the series is \(\sum_{i=0}^{\infty} \frac{2^{i+2}}{i !}\).

Step 2: Simplify the Series

Next, simplify the series. The \(2^{i+2}\) term can also be written as \(4 \cdot 2^i\). The series is now \(\sum_{i=0}^{\infty} \frac{4 \cdot 2^{i}}{i !}\).

Step 3: Final Equivalent Series

The series is therefore equivalent to \(\sum_{i=0}^{\infty} \frac{4 \cdot 2^{i}}{i !}\) with the index of summation beginning at \(n=0\) just as required.

Key Concepts

Index of Summation

Understanding the index of summation is crucial when working with series. It indicates where the summation starts and how it proceeds. In the given exercise, the series initially starts with the index of summation at 'n=2'. If we want to start at 'n=0', we need to adjust our series accordingly. This adjustment is done through a change of index.

Think of the index of summation as the 'counter' in the sum. When you re-index a series, it's akin to resetting the counter's starting point. For the series \( \sum_{n=2}^{\infty} \frac{2^{n}}{(n-2) !} \), we introduced a new index, \( i = n-2 \), which essentially shifts the original series so that it now starts at \( i = 0 \). It preserves the elements of the series but counts them in a new way. Learning to re-index a series accurately is a valuable skill in the study of mathematics and series.

Infinite Series

An infinite series is a sum of infinitely many terms. It's an important concept in calculus and higher mathematics, where it is used to represent functions and solve problems that finite sums can't address. In the series \( \sum_{n=2}^{\infty} \frac{2^{n}}{(n-2) !} \) from our exercise, the upper limit of summation is infinity, indicating that the series goes on indefinitely.

To work with an infinite series, one often needs to find its convergence or divergence. A convergent series approaches a fixed number as more and more terms are added, while a divergent series fails to settle on a single value. This particular series resembles an exponential series, which often converges under certain conditions. Handling infinite series requires careful attention to ensure that manipulations preserve the series' convergence or divergence characteristics.

Factorial Notation

Factorial notation is a fundamental mathematical concept, usually denoted by an exclamation mark (!). It represents the product of all positive integers up to a specified number. For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \). In the context of series and, more specifically, our exercise, factorials often appear in the denominators when dealing with exponential growth or sequences involving terms that increase or decrease rapidly.

In our solution, the original series used \( (n-2)! \) in the denominator. Factoring factorials within a series can be tricky, but it's manageable once you understand the definition and can compute simple factorials. This understanding is critical in simplifying series expressions, as shown by transforming the original series to \( \sum_{i=0}^{\infty} \frac{4 \cdot 2^{i}}{i !} \) while maintaining the series' integrity. This kind of manipulation often helps in analyzing the series further for convergence or in finding a more compact form to work with.