Let's address each part of the question in detail:

a) Simplify the expression

The sequence is not fully provided, but based on the context, I’ll assume that you’re working with an expression of the form:

$\frac{a^k}{k!}$

where $a$ is some fixed integer and $k!$ represents the factorial of $k$. Let's simplify this expression:

$\frac{a^k}{k!}$

This is already in its simplest form. The expression represents the integer $a$ raised to the power $k$ divided by the factorial of $k$, and no further algebraic simplifications can be made. So, the expression is:

$\boxed{\frac{a^k}{k!}}$

b) Write the divisibility statement

We want to express that there exists an integer $a$ such that the expression is an integer for all $k$.

Using the simplified expression from part (a), we can state:

$\exists a \in \mathbb{Z} \text{ such that } \forall k \in \mathbb{N}, \ \frac{a^k}{k!} \in \mathbb{Z}$

This statement means that there exists an integer $a$ such that for every positive integer $k$, the value of $\frac{a^k}{k!}$ is an integer.

c) Explanation: Is this statement true or false?

The statement is false. Here’s why:

1. Factorials grow very quickly, much faster than powers of $a$ as $k$ increases.

2. As $k$ increases, the factorial $k!$ will contain large prime factors, which will not generally divide evenly into $a^k$ unless $a$ is sufficiently large to cancel out these prime factors. However, for any fixed $a$, there will be some $k$ such that the prime factors in $k!$ are greater than those in $a^k$.

For example, if $a = 1$, the expression is simply $\frac{1}{k!}$, which is not an integer for any $k > 1$. If $a = 2$, then for $k = 3$, we get:

$\frac{2^3}{3!} = \frac{8}{6} = \frac{4}{3},$

which is not an integer. This generalizes to higher values of $k$.

Prime Factor Argument

The key observation is that if $p$ is the largest prime factor of $k!$, then $p$ must also divide $a^k$ in order for $\frac{a^k}{k!}$ to be an integer. However, for sufficiently large $k$, the prime factors of $k!$ exceed those of $a^k$, making divisibility impossible. Therefore, for any fixed $a$, there will always be some $k$ for which $\frac{a^k}{k!}$ is not an integer.

Thus, the statement is false because no such integer $a$ exists that satisfies the condition for all $k$.

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Let me know if you want further details or have any questions!

5 related questions:

1. What is the growth rate comparison between $a^k$ and $k!$?
2. How can we use prime factorization to analyze divisibility in more complex expressions?
3. What is an example of a sequence where $\frac{a^k}{k!}$ does produce an integer for some values of $k$?
4. How does the factorial function influence the divisibility of expressions involving powers and products?
5. Can Stirling's approximation help us understand the behavior of $\frac{a^k}{k!}$ for large $k$?

Tip:

When analyzing the divisibility of sequences involving powers and factorials, always consider the growth of prime factors in both terms. Factorials grow quickly, introducing larger and larger primes into the denominator.