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