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

1. Oct 24, 2012

gottfried

1. The problem statement, all variables and given/known data
Let Sn = 1+$\frac{1}{2}$+$\frac{1}{3}$+$\frac{1}{4}$...+$\frac{1}{n}$. Show that |S2n-Sn|$\geq$ $\frac{1}{2}$

2. Relevant equations

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 24, 2012

LCKurtz

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.

: 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.

Last edited: Oct 24, 2012
3. Oct 25, 2012

gottfried

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.