Question

For what integers \(n\) is $$ \frac{1}{n !}>\frac{8^{n}}{(2 n) !} ? $$ Prove your answer by induction.

Short answer

Answer: There are no integer values of n that make the inequality hold true.

Step-by-step solution

Base Case

First, let's check the base case for the smallest integer value, which is n = 1. $$ \frac{1}{1!} > \frac{8^{1}}{(2(1))!} $$ $$ 1 > \frac{8}{2!} $$ After evaluating, we get: $$ 1 > \frac{8}{2} $$ Which simplifies to: $$ 1 > 4 $$ Since the above inequality is false, the base case does not hold.

Inductive Step

Now, let's assume the inequality holds true for n = k (k ∈ N) where k>1, which means: $$ \frac{1}{k !} > \frac{8^{k}}{(2k) !} $$ We need to show that the inequality also holds true for n = k+1: $$ \frac{1}{(k + 1)!} > \frac{8^{k + 1}}{(2(k + 1))!} $$

Prove the inductive step

Now we use our assumption to try to prove the inequality for n=k+1. Dividing both sides of our assumption by \((k+1)\) and multiplying them by \(8\) , we get: $$ \frac{8}{(k + 1)(k!)} > \frac{8^{k + 1}}{(2k + 2)!} $$ Now, we need to show that: $$ \frac{1}{(k + 1)!} > \frac{8^{k + 1}}{(2(k + 1))!} $$ Notice that, since \(k>1\), the expression \(\frac{8}{(k+1)}\) is smaller than 1. That implies $$ \frac{8}{(k + 1)(k!)} < \frac{1}{k!} $$ So, using our assumption \(\frac{1}{k !} > \frac{8^{k}}{(2k) !}\), we have: $$ \frac{8}{(k + 1)(k!)} < \frac{1}{k!} < \frac{8^{k}}{(2k) !} $$ Thus, $$ \frac{8}{(k + 1)(k!)} > \frac{8^{k}}{(2k) !}(2k+2)(2k+1) = \frac{8^{k+1}}{(2k+2)!} $$ From that, we can see that the inequality holds true for n=k+1 when it is true for n=k. Since the base case is false we deduce that the inequality is also false for values \(n>1\). In conclusion, there is no integer \(n\) that makes the inequality hold true.

Key concepts

Factorials

Factorials, denoted as \(n!\), represent the product of all positive integers from 1 to \(n\). It is a fundamental concept in combinatorics and provides the foundation for counting permutations and combinations.
Factorials grow rapidly as \(n\) increases, making them useful in calculating large combinatorial quantities. For instance:
\[ \begin{align*} 1! &= 1, \2! &= 2 \times 1 = 2, \3! &= 3 \times 2 \times 1 = 6, \4! &= 4 \times 3 \times 2 \times 1 = 24, \&\vdots \end{align*} \]
This property of rapid growth is critical in the analysis of mathematical problems involving factorials, like the exercise of comparing \(\frac{1}{n!}\) and \(\frac{8^n}{(2n)!}\). The factorial in the denominator of these expressions influences their behavior significantly, making it crucial to understand when tackling problems involving large values of \(n\).

Inequalities

Inequalities are mathematical statements that express the relative size or order of two numbers or expressions. In the context of the exercise, our inequality is \(\frac{1}{n!} > \frac{8^n}{(2n)!}\).
Understanding inequalities involves:

• Comparing Sizes: We evaluate whether one expression truly exceeds or falls below the other.
• Manipulation: Techniques such as cross-multiplication and substitution can help analyze complex inequalities.

This inequality involves factorials, where both sides of the equation diminish as \(n\) increases due to the increasing factorials.
Ultimately, practicing inequalities aids in honing logical thinking skills, essential for proofs by induction and other mathematical reasoning tasks. By systematically proving or disproving inequalities, we develop a deeper understanding of mathematical relationships.

Integer Solutions

Integer solutions refer to solutions of equations or inequalities where variables take on integer values only. In problems like the given exercise, identifying potential integer solutions is crucial. When analyzing \(\frac{1}{n!} > \frac{8^n}{(2n)!}\), it is necessary to check integer values of \(n\) for which this inequality may hold.
Several strategies can be applied when seeking integer solutions:

• Substitution: Plugging in successive integer values to check validity of the inequality.
• Pattern Recognition: Noticing if certain integers might fit a solution based on observed patterns.
• Logical Reasoning: Employing principles, like induction, to ascertain which integers might be included or excluded as solutions.

In the exercise, the base case showed no integer \(n\) satisfies the inequality, thus illustrating the absence of integer solutions. Understanding integer solutions can substantially aid in problem-solving, providing clarity about the feasible values for variables within a given constraint.