The given expression is:

$\frac{(-1)^n \cdot 4}{n^{1.1}}$

This expression represents a sequence or function depending on the value of $n$. Here's how it behaves for different $n$:

1. The term $(-1)^n$ alternates the sign:

   • For even $n$, $(-1)^n = 1$.
   • For odd $n$, $(-1)^n = -1$.

2. The denominator $n^{1.1}$ grows as $n$ increases because it is raised to the power of 1.1, which is slightly more than 1.

So, this expression alternates between positive and negative values due to $(-1)^n$, and its magnitude decreases as $n$ increases due to the denominator $n^{1.1}$.

Example Evaluations:

For specific values of $n$:

• $n = 1$: $\frac{(-1)^1 \cdot 4}{1^{1.1}} = \frac{-4}{1} = -4$
• $n = 2$: $\frac{(-1)^2 \cdot 4}{2^{1.1}} = \frac{4}{2^{1.1}} \approx \frac{4}{2.14} \approx 1.87$

As $n \to \infty$, the expression approaches 0 because $n^{1.1}$ grows faster than the constant 4.

Would you like further details or have any specific questions about this expression?

Here are five related questions:

1. How does the growth rate of $n^{1.1}$ compare with other powers of $n$?
2. What is the behavior of this expression as $n$ approaches infinity?
3. How would the expression change if the exponent on $n$ was different (e.g., $n^{1.5}$)?
4. Can this expression be integrated over a certain interval of $n$?
5. What would be the behavior if $(-1)^n$ were replaced by a different periodic function?

Tip: Alternating sign terms like $(-1)^n$ are common in alternating series. Keep this in mind when analyzing the convergence of such expressions.