Prove that the harmonic series is divergent

The integral test works well for that one.

 

Cauchy condensation, sum(a_n) converges <=> sum(2^k*a_(2k)) converges, only works for monotonically decreasing a_n though

 

For Cauchy condensation?? You use log?? You can prove that test true by the comparison test. No logs required

 

[tex]
H = \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...
[/tex]

Notice that 1/3 + 1/4 > = 1/4 + 1/4 = 1/2.

And that that 1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2

So, in essence,

[tex]

H > 1 + \sum_{n=1}^{\infty} \frac{1}{2}
[/tex]

The latter part diverges, of course.

EDIT: This is the intuitive way. There is also an elementary, but rigorous way of doing it without the integral test.

 

You pick one of the a_2^k terms and since it's decreasing,
2^k*a_2^k
<a_(2^k)+...
+a_(2^(k+1)-1) , by summing over these you Get series comparison