Question

Determine the radius of convergence of the given power series. $$\sum _{n=1}^{\mathrm{\infty }}\frac{n!x^n}{n^n}$$

Short answer

Answer: The radius of convergence of the given power series is $e$.

Step-by-step solution

Write down the general term of the power series

The given power series is: $$\sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{n!x^n}{n^n}$$ where the general term is: $$a_n=\frac{n!x^n}{n^n}$$

Apply the Ratio Test

Using the Ratio Test, we need to find the limit of the ratio of consecutive terms as n approaches infinity: $$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\frac{(n+1)!x^{n+1}}{(n+1{)}^{n+1}}}{\frac{n!x^n}{n^n}}\right|$$

Simplify the expression

We can simplify this expression by canceling out common factors and using properties of exponents: $$=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{(n+1)!x^{n+1}{n}^n}{(n+1{)}^{n+1}n!x^n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{(n+1)x}{(n+1)}\frac{n^n}{(n+1{)}^n}\right|$$

Take the limit as n approaches infinity

Now, we need to compute the limit as n approaches infinity: $$=\underset{n\to \mathrm{\infty }}{lim}\left|x\frac{n^n}{(n+1{)}^n}\right|$$ Use L'Hopital's rule for the limit, taking into account that the limits inside the expression tend to be of the form $0/0$: $$=\left|x\underset{n\to \mathrm{\infty }}{lim}\frac{n^n}{(n+1{)}^n}\right|=x\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n}{n+1}\right)}^n$$ Using the fact that $\underset{n\to \mathrm{\infty }}{lim}(1+\frac{a}{n}{)}^n=e^a$, we have: $$=x\underset{n\to \mathrm{\infty }}{lim}{(1-\frac{1}{n+1})}^{-n}=x\frac{1}{e}.$$

Determine the radius of convergence

For the series to converge, the limit found in Step 4 must be less than 1: $$|x\frac{1}{e}|<1$$ So, the radius of convergence (R) is given by: $$R=\frac{1}{\frac{1}{e}}=e$$ Therefore, the radius of convergence of the given power series is e.

Key concepts

Power Series

A power series is a series in the form of $$\sum _{n=0}^{\mathrm{\infty }}{a}_n(x-c{)}^n$$, where $a_n$ are the coefficients, $x$ is the variable, and $c$ is the center of the series.

The series sums an infinite number of terms with increasing powers of $x$, and the behavior of these terms as $n$ grows large determines whether the series converges or diverges for different values of $x$. Power series can represent functions and are useful in calculus because they allow us to handle functions that might otherwise be difficult to work with.

A particularly important aspect of power series is their radius of convergence. This is the distance from the center $c$ within which the series converges. Outside of this radius, the series diverges. Finding the radius of convergence is critical, as it tells us the interval in which the series truly represents the function it's associated with.

Understanding the concept of power series helps in numerous areas of mathematics and engineering, such as solving differential equations and analyzing complex functions.

Ratio Test

The Ratio Test is a method used to determine the absolute convergence of an infinite series.

When given a series $\sum a_n$, the Ratio Test involves looking at the limit of the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ as $n$ approaches infinity. If this limit is less than 1, then the series is absolutely convergent. If it's greater than 1, the series diverges, and if it equals 1, the test is inconclusive.

Applying the Ratio Test

To apply the test, we compute $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|$. For the power series in the exercise, the Ratio Test simplified to $x\underset{n\to \mathrm{\infty }}{lim}(\frac{n}{n+1}{)}^n$.

This test is particularly useful for series with factorials, exponential functions, or powers of $n$, as these often result in a ratio that simplifies to a meaningful limit. In our example, we notice how neatly the terms reduce, allowing us to easily compare the limit to 1 and thereby determine the series' behavior.

L'Hopital's Rule

L'Hopital's rule is an invaluable technique for evaluating limits that result in indeterminate forms, like $0/0$ or $\mathrm{\infty }/\mathrm{\infty }$.

This rule states that if we have a limit of the form $\underset{x\to c}{lim}\frac{f(x)}{g(x)}$ and both $f(c)$ and $g(c)$ are 0 or both tend to infinity, then the limit can be found by taking the derivative of the numerator and the denominator separately and then taking their limit of the ratio.

Using L'Hopital's Rule

In our exercise, while L'Hopital's rule wasn't explicitly used in the solutions, it is typically applied in cases where simplification leads to an indeterminate form. We instead used a property of exponential functions for our limit, $\underset{n\to \mathrm{\infty }}{lim}(1+\frac{a}{n}{)}^n=e^a$, to find the value of the expression as $n$ approaches infinity. In more complex scenarios where such direct application is not possible, L'Hopital's rule would be instrumental in finding the limit and thus the radius of convergence for the power series.