Question

As in Example 1, use the ratio test to find the radius of convergence \(R\) for the given power series. $$ \sum_{n=0}^{\infty} n !(t-1)^{n} $$

Short answer

Answer: The radius of convergence for the given power series is R = 0.

Step-by-step solution

Write the general term of the series

The given power series can be written as the following: $$ a_n = n !(t-1)^{n} $$

Write the general term a_{n+1}

The next term of the power series can be written as: $$ a_{n+1} = (n+1)!(t-1)^{n+1} $$

Set up the ratio test

The ratio test can be described as the following limit: $$ L = \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} $$

Substitute a_n and a_{n+1} into the ratio test

Plug the general terms a_n and a_{n+1} into the ratio test: $$ L = \lim_{n \to \infty} \frac{|(n+1)!(t-1)^{n+1}|}{|n !(t-1)^{n}|} $$

Simplify the limit expression

We can simplify the limit expression by canceling out some common terms: $$ L = \lim_{n \to \infty} \frac{(n+1)(t-1)}{1} $$

Take the limit

Now, let's take the limit as n approaches infinity: $$ L = (t-1) \lim_{n \to \infty} (n+1) $$ Since the limit goes to infinity, the ratio test is inconclusive for this series.

Determining the radius of convergence

Since the limit of the ratio test goes to infinity, the power series does not converge, meaning the radius of convergence is R = 0.

Key concepts

Ratio Test

The Ratio Test is a powerful method used to determine the convergence or divergence of an infinite series, particularly useful for series involved in complex expressions. It usually involves the formation of a ratio between consecutive terms of the series, specifically comparing how terms behave as they move towards infinity. For any given power series \( \sum a_{n}x^{n} \), the test determines the Limit \( L \) as follows:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| \]Here's what to look for when applying the Ratio Test:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

In the original exercise, the Ratio Test revealed that \( L = (t-1) \lim_{n \to \infty} (n+1) \). Because this limit reached infinity, which implies \( L > 1 \), the series diverges absolutely, meaning there is no radius of convergence other than zero.

Power Series

A Power Series is a series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where \( a_n \) and \( c \) are constants. The variable \( x \) can take different values, and the series can converge or diverge depending on its structure and the value of \( x \). Power Series are used in various areas of mathematics including calculus, differential equations, and even complex analysis. They are extremely useful in representing functions as infinite sums, especially functions that are not easily integrable or differentiable by other means. The series in the original problem \( \sum_{n=0}^{\infty} n!(t-1)^n \) is centered around \( x = 1 \). The Ratio Test concluded a radius of zero because the factorial term \( n! \) grows too quickly when compared to the polynomial term \((t-1)^n \), leading the series to diverge for any \( t eq 1 \).

Factorials in Series

Factorials often appear in sequences and series, causing significant growth in the terms as \( n \) becomes large. Notationally, \( n! \) ("n factorial") is the product of all positive integers up to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).Their presence in a series like \( \sum_{n=0}^{\infty} n!(t-1)^n \) poses a challenge because factorial terms, \( n! \), can grow faster than exponential and polynomial terms.In our exercise, the \( (n+1)! \) in \( a_{n+1} \) means that each new term rapidly increases due to the factorial's growth. This leads to divergence in the context of the Ratio Test, showing that irrespective of the base \( t-1 \), the increasing factorial outweighs any potential convergence from \( (t-1)^n \). Therefore, factorials in series often require special attention as they tend to push towards divergence unless otherwise balanced by other elements of the series.