Question

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{(n !)^{3} e^{3 n}}{(3 n) !}$$

Short answer

The series converges.

Step-by-step solution

Identify the terms of the series

The given series is \(\sum_{n=0}^{\infty} \frac{(n !)^{\3} e^{\3n}}{(3n) !}\). Let \(a_n\) be the \(n\)th term of the series, so \(a_n = \frac{(n !)^{\ 3} e^{\3n}}{(3n) !}\).

Apply the ratio test

For the ratio test, we need to find the limit \(\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|\).

Find \(a_{n+1}\)

Calculate the \(n+1\)th term: \(a_{n+1} = \frac{((n+1) !)^{\3} e^{\3(n+1)}}{(3(n+1)) !}\).

Form the ratio

Form the ratio \(\left| \frac{a_{n+1}}{a_n} \right| = \frac{((n+1) !)^{\3} e^{\3(n+1)}}{(3(n+1)) !} \frac{(3n) !}{(n !)^{\3} e^{\3n}}\).

Simplify the ratio

Simplify the ratio step-by-step: \(\left| \frac{((n+1) !)^{\3} e^{\3(n+1)}}{(3(n+1)) !} \frac{(3n) !}{(n !)^{\3} e^{\3n}} \right| = \frac{((n+1) (n!)^{\3}) \times (n+1)^2 \times e^{\3n+3}}{(3n+3)(3n+2)(3n+1) (3n)!} \times \frac{(3n)!}{(n!)^3 e^{\3n}}\).

Further simplification

Continue simplifying: \(\left| \frac{((n+1)^3 (n!)^{\3} e^{\3n} e^{\3})}{(3n+3)(3n+2)(3n+1) (n!)^{\3} (3n)!} \times \frac{(3n) !}{(n !)^{\3} e^{\3n}} \right| = \frac{e^{\3} (n+1)^3}{(3n+3)(3n+2)(3n+1)}\).

Find the limit

Take the limit as \(n \to \infty\): \(\lim _{{n \to \infty}} \left | \frac{e^{\3} (n+1)^3}{(3n+3)(3n+2)(3n+1)} \right |\). Since the highest degree terms dominate, this simplifies to \( \frac{e^{\3} n^3}{27 n^3} = \frac{e^{\3}}{27} = \frac{e^{\3}}{3^3} \).

Conclusion of the ratio test

If the limit \(\lim _{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| < 1\), the series converges. Here, \( \frac{e^{\3}}{27} < 1\). Therefore, the series converges.

Key concepts

ratio test

The ratio test is a convenient tool to determine whether an infinite series converges or diverges. It examines the ratio of successive terms in the series to identify a specific limit. If this limit is less than 1, the series converges; if greater than 1, the series diverges. If the limit is exactly 1, the test is inconclusive.

Here's how the ratio test works step-by-step:

• Identify the general term of the series, typically denoted as \(a_n\).
• Calculate the ratio of the \(n+1\)-th term to the \(n\)-th term: \left| \frac{a_{n+1}}{a_n} \right|\.

After simplifying this ratio, take the limit as \ approaches infinity. Symbolically,
\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|. \]
If the limit is less than 1, the series is **convergent**. If greater than 1, it is **divergent**. If equal to 1, the test does not help us conclude the behavior of the series.

series convergence

Understanding series convergence is crucial in calculus and higher mathematics. When we talk about a series converging, we mean that the sequence of partial sums of the series tends to a definite value as the number of terms increases.

If a series converges, we can say:
\[\sum_{n=1}^{\infty} a_n = L,\]
where \(L\) is a finite number.

• Convergent Series: A series \(\sum a_n\) converges if the limit \(\lim_{n \to \infty} S_n = \sum_{n=1}^{\infty} a_n \) exists and is finite.
• Divergent Series: If the limit does not exist or is infinite, then the series diverges.

Various tests, such as the ratio test, root test, and integral test, help determine the convergence or divergence of a series. Specifically, the ratio test is particularly useful with factorials and exponential functions, as it simplifies the process of identifying the series' behavior.

factorials

Factorials are fundamental in many branches of mathematics and play a pivotal role in series and sequences. Represented by the symbol \(!\), a factorial of a non-negative integer n (denoted as \(n!\)) is the product of all positive integers less than or equal to \(n\). For instance, $$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120.$$
Some key properties of factorials are:

• The factorial of 0 is 1: \( 0! = 1 \).
• For any positive integer \(n\), \(n! = n \cdot (n-1)!\).
• Factorials grow extremely quickly as \(n\) increases.

In the context of series and the ratio test, handling factorials can be simplified using approximations like Stirling's approximation. This says for large \(n\), \(n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n\), which aids in understanding the behavior of terms involving factorials for very large \(n\).