Proving S2n-Sn is greater than a half in terms of the harmonic series

gottfried
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Homework Statement


Let Sn = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{n}. Show that |S2n-Sn|\geq \frac{1}{2}


Homework Equations





The Attempt at a Solution



So I'm going to try and use induction

Base case let n=1

|S2n-Sn| = \frac{1}{2}

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|\geq \frac{1}{2} and S2m+2-S2m = \frac{1}{2m+1}+\frac{1}{2m+2}> 0

Therefore |S2m+2-Sm|= |S2m+2-S2m+S2m-S2m+2|> \frac{1}{2}

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?
 
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gottfried said:

Homework Statement


Let Sn = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{n}. Show that |S2n-Sn|\geq \frac{1}{2}


Homework Equations





The Attempt at a Solution



So I'm going to try and use induction

Base case let n=1

|S2n-Sn| = \frac{1}{2}

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|\geq \frac{1}{2} and S2m+2-S2m = \frac{1}{2m+1}+\frac{1}{2m+2}> 0
I don't quite follow that last step. But since ##\{S_n\}## is an increasing sequence you have$$
|(S_{2m+2}-S_{2m}) + (S_{2m}-S_{m+1})|=(S_{2m+2}-S_{2m}) + (S_{2m}-S_{m+1})\ge \frac 1 2$$since the quantities in parentheses are positive.
[Edit]: Ignore the highlighted paragraph. I wasn't completely awake. Look below:

But this problem hardly needs induction. Write out ##S_{2n}-S_n##, look at how many terms there are, and underestimate each by the smallest one.
 
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So there are n terms between S2n and Sn the smallest of which is \frac{1}{2n}

It follows that |S2n - Sn|>n(\frac{1}{2n} ) =\frac{1}{2}

That is frustratingly simple. I feel silly but thanks a lot.
 
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