Question 465619
It is quite difficult to come up with an explicit formula for the nth partial sum of the harmonic series, but there is a famous theorem which says that


*[tex \LARGE \lim_{n \to \infty} (\sum_{i=1}^{n} \frac{1}{i} - \ln{n}) = \gamma]


This essentially says that the difference between the nth partial sum and ln n approaches a constant (known as the Euler-Mascheroni constant) as n gets really large.



Obviously, the infinite sum


*[tex \LARGE \sum_{i=1}^{\infty} \frac{1}{i}]


diverges due to the integral test. Since


*[tex \LARGE \int_{1}^{\infty} \ln (x)\, dx = \lim_{x \to \infty} \ln (x) - ln (0) \to \infty]


the integral diverges and so does the infinite sum.


Here is some more information on the nth partial sums of the harmonic series:
http://en.wikipedia.org/wiki/Harmonic_number