Question

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{n !}{(2 n) !}$$

Short answer

The series converges.

Step-by-step solution

Write the General Term of the Series

Given the series \(\sum_{n=0}^{\infty} \frac{n !}{(2 n) !}\), identify the general term \(a_n\) as \(\frac{n !}{(2 n) !}\).

Setup the Ratio Test

The ratio test requires calculating the limit \(\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\). Substitute \(a_n\) and \(a_{n+1}\) into this limit.

Compute \(a_{n+1}\)

First, find the \(a_{n + 1}\) term: \(a_{n + 1} = \frac{(n + 1)!}{(2(n + 1))!} = \frac{(n + 1)!}{(2n + 2)!}\).

Form the Ratio \(\frac{a_{n+1}}{a_n}\)

Now form the ratio: \(\frac{a_{n+1}}{a_n} = \frac{\frac{(n + 1)!}{(2n + 2)!}}{\frac{n!}{(2n)!}} = \frac{(n + 1)!}{(2n+2)!} \cdot \frac{(2n)!}{n!}\).

Simplify the Ratio

Simplify \(\frac{a_{n+1}}{a_n}\): \[\frac{(n+1)}{(2n+2)(2n+1)} = \frac{(n+1)}{((2n+2)!) / ((2n)!)} = \frac{n + 1}{4n^2 + 6n + 2}\]

Find the Limit

Now find the limit:\( \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \rightarrow \infty} \frac{n + 1}{(2n + 2)(2n + 1)} = \lim_{n \rightarrow \infty} \frac{n + 1}{4n^2 + 6n + 2} = 0\).

Conclusion from Ratio Test

Since \(\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1\), by the Ratio Test, the series \(\sum_{n=0}^{\infty} \frac{n !}{(2 n) !}\) converges.

Key concepts

ratio test

The ratio test is a popular method for determining the convergence or divergence of an infinite series. This test relies on calculating the limit of the ratio of consecutive terms in the series. To use the ratio test, follow these steps:

• Identify the general term of the series, often denoted as \(a_n\).
• Compute the next term, \(a_{n+1}\).

Next, form the ratio of these terms, \(\frac{a_{n+1}}{a_n}\), and simplify it. Finally, find the limit of this ratio as \(n\) approaches infinity: \(\text{Limit} = \lim_{n \to \infty}\frac{a_{n+1}}{a_n}\).
The series converges if the limit is less than 1. If the limit equals 1, the test is inconclusive, and if it is more than 1, the series diverges.

convergence of series

Understanding the convergence of series is crucial in mathematics. A series \(\sum_{n=0}^{\infty} a_n\) converges if the sequence of its partial sums converges to a finite limit as \(n\) approaches infinity. In simpler terms, the infinite sum of the terms in the series gets closer and closer to a specific number.The ratio test helps make this determination easily for many types of series. If, after applying the ratio test, the result is a limit less than 1, it means the terms of the series get smaller fast enough to sum up to a finite number.

• For instance, a geometric series \(\sum_{n=0}^{\infty} ar^n\) converges if \(|r|<1\).

Applying these concepts to practice repeatedly helps solidify the understanding of convergence and divergence in series.

factorials in series

Factorials play a significant role in many series, particularly when applying the ratio test. A factorial, denoted as \(n!\), is the product of all positive integers up to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). When factorials are present in series, they often simplify the process of applying the ratio test. Consider the series, \sum_{n=0}^{\infty} \frac{n!}{(2n)!}\, from our initial exercise:

• Identify \(a_n = \frac{n!}{(2n)!}\)
• Compute \(a_{n+1} = \frac{(n+1)!}{(2(n+1))!} = \frac{(n+1)!}{(2n+2)!}\)

These steps show how factorial terms simplify, eventually aiding in determining the convergence of the series.