1. proof of harmonic series

Find an proof for the harmonic series not using the integral test with the following hint.

$\displaystyle \sum_{k = 1}^{\infty} = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) + ...$

observe that

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

$\displaystyle \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} > \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8} = \frac{1}{2}$

It looks like we are comparing grouping of the harmonic series to another series. Each grouping is always grater than 1/2 so the harmonic series is greater than $\displaystyle 1 + \sum_{k = 1}^{\infty} \frac{1}{2}$ which diverges.

$\displaystyle 1 + \sum_{k = 1}^{\infty} \frac{1}{2} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...$

Therefor the harmonic series diverges.

2. Originally Posted by diroga
Find an proof for the harmonic series not using the integral test with the following hint.

$\displaystyle \sum_{k = 1}^{\infty} = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) + ...$

observe that

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

$\displaystyle \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} > \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8} = \frac{1}{2}$

It looks like we are comparing grouping of the harmonic series to another series. Each grouping is always grater than 1/2 so the harmonic series is greater than $\displaystyle 1 + \sum_{k = 1}^{\infty} \frac{1}{2}$ which diverges.

$\displaystyle 1 + \sum_{k = 1}^{\infty} \frac{1}{2} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...$

Therefor the harmonic series diverges.
That is correct.
You may be a bit more formal. But the idea is correct.

3. is using $\displaystyle 1 + \sum_{k = 1}^{\infty} \frac{1}{2} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...$ an acceptable comparison