Question

Determine the convergence of the given series. State the test used; more than one test may be appropriate. $$\sum_{n=1}^{\infty} \frac{n^{4} 4^{n}}{n !}$$

Short answer

The series converges by the Ratio Test.

Step-by-step solution

Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{n^{4} 4^{n}}{n!} \). We need to check if this series converges or diverges.

Consider the Ratio Test

The Ratio Test is useful for series with factorials. For the series \( \sum_{n=1}^{\infty} a_n \), the Ratio Test states that if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), then the series converges. Let's apply it to our series.

Apply the Ratio Test

Compute \( a_n = \frac{n^4 4^n}{n!} \) and \( a_{n+1} = \frac{(n+1)^4 4^{n+1}}{(n+1)!} \). The ratio is:\[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^4 4^{n+1} n!}{n^4 4^n (n+1)!} = \frac{(n+1)^4 4}{n^4 (n+1)} = \frac{(n+1)^3 4}{n^4} \]

Simplify and Find the Limit

We simplify \( \frac{(n+1)^3 4}{n^4} \) as \( \frac{4(n+1)^3}{n^4} \). Expand \( (n+1)^3 \):\[ (n+1)^3 = n^3 + 3n^2 + 3n + 1 \]So, the expression becomes:\[ \frac{4(n^3 + 3n^2 + 3n + 1)}{n^4} = 4 \left( \frac{n^3}{n^4} + \frac{3n^2}{n^4} + \frac{3n}{n^4} + \frac{1}{n^4} \right) \]Simplifying each term:\[ = 4 \left( \frac{1}{n} + \frac{3}{n^2} + \frac{3}{n^3} + \frac{1}{n^4} \right) \]

Evaluate the Limit

As \( n \to \infty \), \( \frac{1}{n} \to 0 \), \( \frac{3}{n^2} \to 0 \), \( \frac{3}{n^3} \to 0 \), and \( \frac{1}{n^4} \to 0 \). Therefore:\[ \lim_{n \to \infty} 4 \left( \frac{1}{n} + \frac{3}{n^2} + \frac{3}{n^3} + \frac{1}{n^4} \right) = 0 \]

Conclusion Based on the Ratio Test

Since \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1 \), the Ratio Test confirms that the series \( \sum_{n=1}^{\infty} \frac{n^{4} 4^{n}}{n!} \) converges.

Key concepts

Ratio Test

The Ratio Test is a popular method used to determine the convergence of an infinite series, especially when that series involves factorials or exponential functions. This test works by considering the ratio between successive terms of the series.
To apply the Ratio Test, you first find the absolute value of the ratio of the \( n+1 \)th term to the \( n \)th term of the series. The mathematical expression for this is:

• For a series \( \sum_{n=1}^{\infty} a_n \), compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, or infinite, the series diverges.
• If the limit equals 1, the test is inconclusive.

This test is particularly simple to use when \( a_n \) involves factorial expressions, as these can often simplify nicely as shown in the problem with \( a_n = \frac{n^4 4^n}{n!} \). The main trick is dealing with factorial growth as they tend to simplify ratios of terms.

Factorial Series

Series that involve factorials, like the one in our exercise, often call for strategies tailored to handling the rapid growth of factorial terms. Factorials, denoted as \( n! \), grow incredibly quickly, far outpacing exponential functions like \( 2^n \) or \( 3^n \). Therefore, when a series contains a factorial in the denominator, it already has a built-in tendency towards convergence because the denominator of each term increases rapidly.
For the factorial series \( \sum_{n=1}^{\infty} \frac{n^4 4^n}{n!} \):

• The term \( n^4 \) grows polynomially.
• The term \( 4^n \) grows exponentially.
• The factorial \( n! \) grows faster than either \( n^4 \) or \( 4^n \).

This implies that, as \( n \) becomes very large, the growth of \( n! \) in the denominator is significant enough to outweigh the exponential and polynomial growth in the numerator. This behavior strongly aids in the convergence of the series when assessed using tests like the Ratio Test.

Limit Evaluation

Evaluating limits is a crucial component of the Ratio Test. After substituting successive terms into the ratio, you frequently arrive at complex numerical expressions involving powers or multiples of \( n \). Simplifying these expressions and evaluating their limits as \( n \to \infty \) is key.
For the expression obtained from our series \( \frac{4(n+1)^3}{n^4} \):

• Break down the numerator \( (n+1)^3 = n^3 + 3n^2 + 3n + 1 \).
• Express the entire fraction as \( 4 \left( \frac{n^3}{n^4} + \frac{3n^2}{n^4} + \frac{3n}{n^4} + \frac{1}{n^4} \right) \).
• Simplify to \( 4 \left( \frac{1}{n} + \frac{3}{n^2} + \frac{3}{n^3} + \frac{1}{n^4} \right) \).

As \( n \) approaches infinity, each component of the expression approaches zero. Thus, the limit of the whole expression is zero. This confirms the absolute convergence of our series using the Ratio Test, as the limit \(< 1\). Understanding how to evaluate these limits accurately ensures reliable conclusions about the series' behavior.