The question in the image is about a sequence of numbers defined by $a_n = \left( -\frac{1}{2} \right)^n$, and the task is to determine from which $n$ the absolute value of this sequence becomes and stays less than 0.001, or more generally, less than some given positive $\epsilon$.

Let's break down the problem:

1. The general form of the sequence is $a_n = \left( -\frac{1}{2} \right)^n$.

   • When $n = 0$, $a_0 = \left( -\frac{1}{2} \right)^0 = 1$.
   • When $n = 1$, $a_1 = \left( -\frac{1}{2} \right)^1 = -\frac{1}{2}$.
   • When $n = 2$, $a_2 = \left( -\frac{1}{2} \right)^2 = \frac{1}{4}$.
   • As $n$ increases, the values alternate in sign, but the magnitude decreases as $\frac{1}{2^n}$.

2. We are tasked to find when the absolute value of the sequence $|a_n|$ becomes less than 0.001. That means: $\left| \left( -\frac{1}{2} \right)^n \right| = \left( \frac{1}{2} \right)^n < 0.001$

3. Solving this inequality: $\left( \frac{1}{2} \right)^n < 0.001$ Taking the logarithm on both sides (base 10): $n \log_{10} \left( \frac{1}{2} \right) < \log_{10}(0.001)$ Since $\log_{10} \left( \frac{1}{2} \right) = -\log_{10}(2)$, the equation becomes: $n (-\log_{10}(2)) < -3$ Therefore: $n > \frac{3}{\log_{10}(2)}$ Using $\log_{10}(2) \approx 0.3010$: $n > \frac{3}{0.3010} \approx 9.97$ So, $n \geq 10$.

Thus, starting from $n = 10$, the absolute value of the sequence $a_n$ will be less than 0.001.

Would you like more details or have any further questions?

Related Questions:

1. What is the general behavior of the sequence $a_n = \left( -\frac{1}{2} \right)^n$?
2. How does logarithmic calculation help in solving inequalities?
3. How can we estimate when a sequence converges to a specific value?
4. What are the properties of sequences that alternate in sign but decrease in magnitude?
5. How would this problem change if the base of the exponent were different, e.g., $\left( -\frac{1}{3} \right)^n$?

Tip: When solving exponential inequalities, logarithms are a powerful tool to simplify and solve for unknowns.