Prove that 1 + - + - n2 < 2 — — is tme for all integers n > 1.
[5 Marks]

To prove that \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + (-1)^{n+1} \frac{1}{n} < 2 \) for all integers \( n > 1 \), we can use mathematical induction. 

### Step 1: Define the Series
Let \( S_n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + (-1)^{n+1} \frac{1}{n} \).

We want to show that \( S_n < 2 \) for all integers \( n > 1 \).

### Step 2: Base Case
For \( n = 2 \):
\[
S_2 = 1 - \frac{1}{2} = \frac{1}{2} < 2
\]
The base case holds, as \( S_2 < 2 \).

### Step 3: Inductive Hypothesis
Assume that for some integer \( k \geq 2 \), \( S_k < 2 \) holds true. That is, we assume:
\[
S_k = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + (-1)^{k+1} \frac{1}{k} < 2
\]

### Step 4: Inductive Step
We need to show that \( S_{k+1} < 2 \).

Now, consider \( S_{k+1} \):
\[
S_{k+1} = S_k + (-1)^{k+2} \frac{1}{k+1}
\]

Since \( (-1)^{k+2} \) alternates signs (positive for even \( k+1 \) and negative for odd \( k+1 \)), we add or subtract \( \frac{1}{k+1} \) to \( S_k \).

#### Case Analysis on Addition or Subtraction
1. If \( (-1)^{k+2} \frac{1}{k+1} \) is positive, \( S_{k+1} < 2 \) since \( S_k < 2 \).
2. If \( (-1)^{k+2} \frac{1}{k+1} \) is negative, \( S_{k+1} < 2 \).

### Conclusion
Since both cases satisfy

Prove that 1 - 1/2 + 1/3 - 1/4 + ... + (-1)^(n+1) * (1/n) < 2 for all integers n > 1.

Explore a detailed proof of the inequality 1 - 1/2 + 1/3 - 1/4 + ... + (-1)^(n+1) * (1/n) < 2 for all integers n > 1, using mathematical induction. This advanced high school math problem introduces series, convergence, and induction techniques, suitable for Grades 11-12.

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