What is the greatest value of the positive integer n satisfying the condition \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}?\)
10
Solving the Geometric Series Inequality
The problem asks for the greatest positive integer \(n\) that satisfies a given inequality involving a sum of terms. The sum is \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}}\). This is a finite geometric series.
Identifying the Geometric Series
A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In this case:
- The first term is \(a = 1\).
- The common ratio is \(r = \frac{1/2}{1} = \frac{1/4}{1/2} = \ldots = \frac{1}{2}\).
- The terms are \( \frac{1}{2^0}, \frac{1}{2^1}, \frac{1}{2^2}, \ldots, \frac{1}{2^{n-1}} \).
- There are \(n\) terms in total, from the term with index 0 up to index \(n-1\).
Calculating the Sum of the Series
The sum \(S_n\) of the first \(n\) terms of a geometric series is given by the formula:
\(S_n = a \frac{(1 - r^n)}{(1 - r)}\)
Substituting the values for our series:
\(S_n = 1 \times \frac{(1 - (\frac{1}{2})^n)}{(1 - \frac{1}{2})}\)
\(S_n = \frac{(1 - \frac{1}{2^n})}{\frac{1}{2}}\)
\(S_n = 2 \times (1 - \frac{1}{2^n})\)
\(S_n = 2 - \frac{2}{2^n}\)
\(S_n = 2 - \frac{1}{2^{n-1}}\)
Setting up the Inequality
The given condition is the inequality \(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}\).
Replacing the sum with our formula \(S_n = 2 - \frac{1}{2^{n-1}}\), the inequality becomes:
\(2 - \frac{1}{2^{n-1}} < 2 - \frac{1}{{1000}}\)
Solving the Inequality for n
Now we solve the inequality for the positive integer \(n\). First, subtract 2 from both sides:
\(-\frac{1}{2^{n-1}} < -\frac{1}{{1000}}\)
Multiply both sides by \(-1\). Remember to reverse the inequality sign when multiplying or dividing by a negative number:
\(\frac{1}{2^{n-1}} > \frac{1}{{1000}}\)
Now, take the reciprocal of both sides. Again, remember to reverse the inequality sign:
\(2^{n-1} < 1000\)
We need to find the greatest positive integer \(n\) such that \(2^{n-1}\) is less than 1000. Let's look at powers of 2:
| \(k\) | \(2^k\) |
|---|---|
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
| 4 | 16 |
| 5 | 32 |
| 6 | 64 |
| 7 | 128 |
| 8 | 256 |
| 9 | 512 |
| 10 | 1024 |
We need \(2^{n-1} < 1000\). Looking at the table:
- If \(n-1 = 9\), then \(2^9 = 512\). Since \(512 < 1000\), this is true.
- If \(n-1 = 10\), then \(2^{10} = 1024\). Since \(1024 < 1000\) is false, this is not true.
The greatest integer value for \(n-1\) that satisfies the inequality \(2^{n-1} < 1000\) is 9.
So, we have \(n-1 = 9\). Solving for \(n\):
\(n = 9 + 1\)
\(n = 10\)
Conclusion on the Greatest Positive Integer n
The greatest positive integer \(n\) satisfying the condition \(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}\) is 10.
Revision Table: Key Steps to Solve the Inequality
| Step | Description | Mathematical Expression |
|---|---|---|
| 1 | Identify the sum as a geometric series | \(1 + \frac{1}{2} + \ldots + \frac{1}{2^{n-1}}\) |
| 2 | Find the sum formula for the series | \(S_n = 2 - \frac{1}{2^{n-1}}\) |
| 3 | Substitute sum into the inequality | \(2 - \frac{1}{2^{n-1}} < 2 - \frac{1}{1000}\) |
| 4 | Simplify the inequality | \(\frac{1}{2^{n-1}} > \frac{1}{1000}\) |
| 5 | Rearrange to isolate \(2^{n-1}\) | \(2^{n-1} < 1000\) |
| 6 | Find the largest integer \(k = n-1\) satisfying \(2^k < 1000\) | \(k = 9\) (since \(2^9 = 512\) and \(2^{10} = 1024\)) |
| 7 | Solve for \(n\) | \(n-1 = 9 \Rightarrow n = 10\) |
Additional Information: Understanding Geometric Series and Inequalities
Geometric Series: A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sum of an infinite geometric series with \(|r| < 1\) converges to \(\frac{a}{1-r}\). In this problem, the series is finite.
Inequalities: Mathematical statements comparing two expressions using symbols like < (less than), > (greater than), \(\leq\) (less than or equal to), or \(\geq\) (greater than or equal to). When solving inequalities, multiplying or dividing both sides by a negative number requires reversing the inequality sign.
The series in the problem is a part of the infinite geometric series \(1 + \frac{1}{2} + \frac{1}{4} + \ldots\), which converges to \(\frac{1}{1 - 1/2} = \frac{1}{1/2} = 2\). The inequality \(S_n < 2 - \frac{1}{1000}\) means the sum is very close to, but slightly less than, 2. As \(n\) increases, the sum \(S_n = 2 - \frac{1}{2^{n-1}}\) gets closer and closer to 2. We are looking for the largest \(n\) before \(S_n\) becomes too large (specifically, not less than \(2 - \frac{1}{1000}\)). This happens when \(\frac{1}{2^{n-1}}\) becomes too small (specifically, not greater than \(\frac{1}{1000}\)). This leads to the condition \(2^{n-1} < 1000\).
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