Problem 21 - Single Variable Calculus

Question

Find the radius of convergence and the interval of convergence of the power series. $$ \sum_{n=1}^{\infty} \frac{2^{n}(x+2)^{n}}{n^{n}} $$

Step-by-Step Solution

Answer

The radius of convergence for the given power series is \(\frac{e}{2}\), and the interval of convergence is \(-\frac{e}{2} - 2 \le x \le \frac{e}{2} - 2\).

Step 1: Apply the Ratio Test

Apply the Ratio Test, which states that the series converges if: \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1\) Where, \(a_n = \frac{2^n (x+2)^n}{n^n}\)

Step 2: Determine the Ratio of Consecutive Terms

Find the ratio of consecutive terms, i.e., \(\left|\frac{a_{n+1}}{a_n}\right|\): \[ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\frac{2^{n+1}(x+2)^{n+1}}{(n+1)^{n+1}}}{\frac{2^n (x+2)^n}{n^n}}\right| \]

Step 3: Simplify the Ratio

Simplify the ratio: \[ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{2^{n+1}(x+2)^{n+1}}{(n+1)^{n+1}} \cdot \frac{n^n}{2^n (x+2)^n}\right| \\ \left|\frac{a_{n+1}}{a_n}\right| = \frac{2}{\left|\frac{n+1}{n}\right|^n}\left|x+2\right| \]

Step 4: Determine the Limit

Find the limit as n approaches infinity: \( \lim_{n \to \infty} \frac{2}{\left|\frac{n+1}{n}\right|^n}\left|x+2\right| \) As n approaches infinity, \(\left|\frac{n+1}{n}\right|^n\) approaches \(e\), therefore: \( \lim_{n \to \infty} \frac{2}{\left|\frac{n+1}{n}\right|^n}\left|x+2\right| = \frac{2}{e}|x+2|\) For the series to converge, this limit must be less than 1: \(\frac{2}{e}|x+2| < 1\)

Step 5: Find the Radius of Convergence

Solve the inequality for 'x': \(|x+2| < \frac{e}{2}\) This inequality represents the interval of convergence, and the radius of convergence is equal to half the length of this interval: Radius of Convergence = \(\frac{e}{2}\)

Step 6: Analyze the Endpoints of the Interval

Check the convergence of the series at the endpoints: For x = -\(\frac{e}{2}\) - 2, the series simplifies to: \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^n}\) For x = \(\frac{e}{2}\) - 2, the series simplifies to: \(\sum_{n=1}^{\infty} \frac{(-1)^n 2^{n}}{n^n}\) Both of these series converge by comparison tests. Therefore, the interval of convergence includes the endpoints.

Step 7: Write the Interval of Convergence

The interval of convergence is: \(-\frac{e}{2} - 2 \le x \le \frac{e}{2} - 2\) In conclusion, the radius of convergence of the power series is \(\frac{e}{2}\) and the interval of convergence is \(-\frac{e}{2} - 2 \le x \le \frac{e}{2} - 2\).

Key Concepts

Power Series

A power series is a type of infinite series that is written in the form:

• \(\sum_{n=0}^{\infty} a_n (x-c)^n\)

where \(a_n\) represents the coefficient for each term, \(x\) is the variable, and \(c\) is the center of the series.
The series describes a function that can be compared to a polynomial with infinite terms.
Power series are critical in mathematical analysis due to their ability to represent complex functions.
In the given exercise, the power series is:

• \(\sum_{n=1}^{\infty} \frac{2^n(x+2)^n}{n^n}\)

Notice here that the series is centered at \(-2\), which is indicated by \((x+2)\).
This slight shift makes the series evaluate around \(x = -2\).

Ratio Test

The Ratio Test is a tool used to determine the convergence or divergence of infinite series.
It is particularly useful for power series and is expressed as follows:

• The series \(\sum a_n\) converges if \(\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1\).
• It diverges if the limit is greater than 1.

In this exercise, the Ratio Test was applied to find \(\left|\frac{a_{n+1}}{a_n}\right|\) for our power series, with \(a_n = \frac{2^n (x+2)^n}{n^n}\).
Through simplification, it led us to evaluate the limit:

• \((\lim_{n \to \infty} \frac{2}{\left|\frac{n+1}{n}\right|^n} |x+2| < 1)\)

This step is crucial in determining how the series behaves as \(n\) becomes very large.

Interval of Convergence

The interval of convergence is the set of \(x\)-values for which a power series converges.
After determining the radius of convergence, the interval can be found by checking which \(x\)-values satisfy the convergence criteria.
It involves checking the inequality derived from the Ratio Test, \(|x+2| < \frac{e}{2}\).
In solving this inequality:

• The center \(c = -2\) and radius of convergence \(\frac{e}{2}\) gives us \(-\frac{e}{2} - 2 < x < \frac{e}{2} - 2\).

Analysis of the endpoints is necessary, as sometimes the series might converge at its bounds.
In the exercise, both endpoints were included, resulting in the closed interval:

• \(-\frac{e}{2} - 2 \le x \le \frac{e}{2} - 2\)

Limit of a Sequence

The concept of the limit of a sequence is fundamental in calculus and analysis.
It helps determine the behavior of a sequence as the term number \(n\) tends towards infinity.
In the Ratio Test application, finding \(\lim_{n \to \infty} \frac{2}{\left|\frac{n+1}{n}\right|^n} |x+2|\) is essential to evaluate if the series converges.
This limit uses the principle that as \(n\) approaches infinity, certain expressions have predictable outcomes:

• For instance, \(\left|\frac{n+1}{n}\right|^n\) approaches \(e\), a known mathematical constant.

By substituting this result back into the limit expression, we simplify the convergence check of our power series.
Understanding limits in sequences helps predict long-term behavior in various mathematical contexts.