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Posted

is divergent, but my professor said the proof wasn't in the sillibus. I'm sure anyone here could show me why. Please?

Posted

The proof is fairly straight-forward. You need to take careful note of the following grouping:

 

[math]H = 1 + \underbrace{\frac{1}{2}}_{\geq \frac{1}{2}} + \underbrace{\frac{1}{3} + \frac{1}{4}}_{\geq \frac{1}{2}} + \underbrace{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}_{\geq \frac{1}{2}} + \underbrace{\frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16}}_{\geq \frac{1}{2}} \cdots[/math]

 

From this, you should be able to see that if [math]s_n = \sum_{k=1}^n \frac{1}{k}[/math], then [imath]s_{2n} \geq s_n + \frac{1}{2}[/imath].

 

You can also prove this:

 

[math]s_{2n} = s_n + \sum_{k=n+1}^{2n} \frac{1}{k} \geq s_n + n\frac{1}{2n} = s_n + \frac{1}{2}[/math].

 

Now it's easy to prove that [imath]s_{2^n} \geq 1 + \frac{n}{2}[/imath], since [imath]s_{2^{n+1}} = s_{2\cdot2^n} \geq s_{2^n} + \frac{1}{2} \geq 1 + \frac{n+1}{2}[/imath] by induction.

 

All you have to do now is say that you've found a subsequence [imath]s_{2^n} \to \infty[/imath] as [imath]n\to\infty[/imath], and since [imath]s_n[/imath] is a strictly increasing sequence this means that [imath]s_n\to\infty[/imath]. Hence the Harmonic series diverges.

Posted

Another, more far-reaching approach is to look up the proof of the Integral Test (if it's not in your book then I'm sure you will have no trouble finding it on the internet). Then you can apply the integral test to the summand in the harmonic series and show that it diverges.

 

I say that this approach is more far-reaching because you can apply the Integral Test to many more series than just the harmonic series.

Posted
Another' date=' more far-reaching approach is to look up the proof of the Integral Test (if it's not in your book then I'm sure you will have no trouble finding it on the internet). Then you can apply the integral test to the summand in the harmonic series and show that it diverges.

 

I say that this approach is more far-reaching because you can apply the Integral Test to many more series than just the harmonic series.[/quote']

 

I actually just learned the integral test today, so I see how that proofs works.

 

 

Thank you dave for showing me the light!

Posted

That's no problem :) I thought the integral test would be rather like breaking a nut with a sledgehammer for something as simple as the Harmonic Series. Plus I think this way of proving it is quite nice.

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