Problem 38 - Calculus: One and Several Variables

Question

Determine whether the series converges or diverges. $$1+\frac{1 \cdot 2}{1 \cdot 3}+\frac{1 \cdot 2 \cdot 3}{1 \cdot 3 \cdot 5}+\frac{1 \cdot 2 \cdot 3 \cdot 4}{1 \cdot 3 \cdot 5 \cdot 7}+\dots$$

Step-by-Step Solution

Answer

The given series can be represented by the general term \(a_n = \frac{n!}{\prod_{k=1}^{n}(2k-1)}\). Using partial fraction decomposition, the series telescopes to \(S_n = 1 - \frac{1}{2n-1}\). Since the limit as n approaches infinity is 1 (\(\lim_{n\to\infty} S_n = 1\)), the series converges.

Step 1: Find the general term of the series.

We can see that the series has a pattern. Let's write down the first few terms in the series: $$1, \frac{1 \cdot 2}{1 \cdot 3}, \frac{1 \cdot 2 \cdot 3}{1 \cdot 3 \cdot 5}, \frac{1 \cdot 2 \cdot 3 \cdot 4}{1 \cdot 3 \cdot 5 \cdot 7}$$ For the numerators, we have a sequence of factorials: $$1, 2!, 3!, 4!$$ The denominators appear to be the product of the first "n" odd integers: $$1, 1 \cdot 3, 1 \cdot 3 \cdot 5, 1 \cdot 3 \cdot 5 \cdot 7$$ The general term of the series can be represented as: $$a_n = \frac{n!}{\prod_{k=1}^{n}(2k-1)}$$ Where "n" is the term number and the "Π" symbol represents the product of odd integers.

Step 2: Partial fraction decomposition

Let's represent each term as a partial fraction: $$a_n=\frac{n!}{\prod_{k=1}^{n}(2k-1)} = \frac{(n-1)!}{\prod_{k=1}^{n-1}(2k-1)} - \frac{(n-2)!}{\prod_{k=1}^{n-2}(2k-1)}$$ Notice that we are trying to express the general term as a difference: two fractions with familiar patterns.

Step 3: Check for telescoping

Now that we have the partial fraction decomposition, we can check if the series telescopes: $$S_n = \sum_{k=1}^{n} a_k = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots$$ We can see that many terms cancel out, leaving: $$S_n = 1 - \frac{1}{2n-1}$$

Step 4: Analyze the limit

Now, we will analyze the limit of the partial sum as n approaches infinity: $$\lim_{n\to\infty} S_n = \lim_{n\to\infty} \left(1 - \frac{1}{2n-1}\right)$$ Since the limit exists and is equal to 1: $$\lim_{n\to\infty} S_n = 1$$

Step 5: Determine convergence

Since the limit of the partial sum is 1 as n approaches infinity, we can conclude that the series converges. Our final answer is that the given series converges.

Key Concepts

factorials in series

When dealing with series, factorials often appear in the numerator or denominator of terms. A factorial, denoted by an exclamation mark (for example, \(n!\)), means the product of all positive integers up to a given number \(n\). This concept is powerful when analyzing sequences within series.

In the example series given in the exercise, factorials are used in the numerators: \(1, 2!, 3!, 4!\), and so on. Here, factorials are used to systematically increase the product as you progress through each term. This is a key indicator that the sequence grows in a predictable pattern, making it possible to apply analytic methods to determine whether the series converges or diverges.

Understanding how to manipulate these factorial expressions is a gateway to analyzing more complex series problems. Skills here include breaking down terms and leveraging factorial properties to simplify expressions. Keep in mind that as the terms grow larger, the factorial functions tend to grow rapidly, which can heavily influence whether a series converges or not.

telescoping series

A telescoping series is a type of series where many terms cancel each other out, making it easier to determine the sum. Visualize a telescope that retracts or folds into itself; similarly, these series simplify by collapsing terms through cancellation.

In the exercise, after partial fraction decomposition, the series transforms into a telescoping series with terms such as \((1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + \dots\). As you sum these terms, observe the cancellation between consecutive terms, which reduces the series to a more manageable form, often leaving just the first and the last terms.

This characteristic is valuable in ensuring easier convergence analysis. Once it is reduced, the remaining terms define the series' character more clearly. The task is then to evaluate the terms that do not cancel out, providing a simpler path to finding the series' sum or determining its convergence.

partial fraction decomposition

Partial fraction decomposition involves breaking down a complicated fraction into simpler parts, usually because those parts telescope easily or have other properties that make the series easier to sum.

The strategy is crucial in the given exercise. By expressing \(a_n\) as \(\frac{(n-1)!}{\prod_{k=1}^{n-1}(2k-1)} - \frac{(n-2)!}{\prod_{k=1}^{n-2}(2k-1)}\), it becomes apparent that each term is a difference between two fractions. This setup allows the series to telescope, leading to significant simplification.

The decomposition makes the sum of the series more tangible as these smaller parts are more straightforward to manage and often lead to greater cancellation. Learning this method is powerful for solving complex series not just in theoretical exercises but also in applied mathematical contexts.