- #1
gottfried
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Homework Statement
Let Sn = 1+[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{3}[/itex]+[itex]\frac{1}{4}[/itex]...+[itex]\frac{1}{n}[/itex]. Show that |S2n-Sn|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex]
Homework Equations
The Attempt at a Solution
So I'm going to try and use induction
Base case let n=1
|S2n-Sn| = [itex]\frac{1}{2}[/itex]
So true for base case
Assume true for case that n=m
So if n=m+1
|S2m+2-Sm+1|= |S2m+2-S2m+S2m-Sm+1|
Our induction assumption tells us that |S2m-S2m+2|[itex]\geq[/itex] [itex]\frac{1}{2}[/itex] and S2m+2-S2m = [itex]\frac{1}{2m+1}[/itex]+[itex]\frac{1}{2m+2}[/itex]> 0
Therefore |S2m+2-Sm|= |S2m+2-S2m+S2m-S2m+2|> [itex]\frac{1}{2}[/itex]
Have I used induction properly? Since I haven't used the base case at all the proof feels a little like I've made the assumption that it is true and then shown that it is true but at the same time I feel like I haven't broken any of the rules of induction. I'm I wrong?