# Thread: Help with solving a proof by induction

1. ## Help with solving a proof by induction

Problem: Prove that $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} > \frac{2n}{n + 1}$ where $\displaystyle n \geq 2$

What I have so far:

1) Prove for $\displaystyle n = 2$

$\displaystyle 1 + \frac{1}{2} > \frac{2(2)}{2 + 1}$

$\displaystyle 1\frac{1}{2} > \frac{4}{3}$

$\displaystyle 1.5 > 1.333...$ which is true.

2) Assume for $\displaystyle k$

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} > \frac{2k}{k + 1}$ assumed true

3) Assume for $\displaystyle k + 1$

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} + \frac{1}{k + 1} > \frac{2(k + 1)}{(k + 1)+1}$ not sure what to do

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How can I continue proving this, because this is where I get stuck on actually proving the inequality is true via induction.

Also, have I made any mistakes so far? If so, then what and how could I fix them.

Thanks for any help, BG

2. Originally Posted by BG5965
Problem: Prove that $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} > \frac{2n}{n + 1}$ where $\displaystyle n \geq 2$

What I have so far:

1) Prove for $\displaystyle n = 2$

$\displaystyle 1 + \frac{1}{2} > \frac{2(2)}{2 + 1}$

$\displaystyle 1\frac{1}{2} > \frac{4}{3}$

$\displaystyle 1.5 > 1.333...$ which is true.

2) Assume for $\displaystyle k$

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} > \frac{2k}{k + 1}$ assumed true

3) Assume for $\displaystyle k + 1$

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} + \frac{1}{k + 1} > \frac{2(k + 1)}{(k + 1)+1}$ not sure what to do

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How can I continue proving this, because this is where I get stuck on actually proving the inequality is true via induction.

Also, have I made any mistakes so far? If so, then what and how could I fix them.

Thanks for any help, BG
Having assumed for k, you have:

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} > \frac{2k}{k + 1}$

Then you need to prove for k+1 (not "assume"). What you can try is:

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{k} + \frac{1}{k + 1} > \frac{2k}{k + 1} + \frac 1 {k+1} = \frac{2k + 1}{k + 1}$

by using the induction hypothesis.

Then "all you have to do" is prove that:
$\displaystyle \frac{2k + 1}{k + 1} \ge \frac{2(k + 1)}{k + 2}$

Not that I have a clue as to where to begin on that last bit at the moment, probably straightforward but I haven't had enough caffeine.

3. $\displaystyle \frac{2k+1}{k+1} = \frac{(k+1) + k}{k+1} = \frac{k+1}{k+1} + \frac{k}{k+1} = 1 + \frac{k}{k+1}$

$\displaystyle \frac{2(k+1)}{k+2} = \frac{2k+2}{k+2} = \frac{(k+2)+k}{k+2} = \frac{k+2}{k+2} + \frac{k}{k+2} = 1 + \frac{k}{k+2}$

So we have $\displaystyle \frac{2k+1}{k+1} = 1 + \frac{k}{k+1} \geq 1 + \frac{k}{k+2} = \frac{2(k+1)}{k+2}$

And we're done.

Also, if you're having trouble understanding how proofs by induction work, you might want to read this: http://en.wikipedia.org/wiki/Mathematical_induction

4. Thanks so much both of you, I finally get it now!