This series diverges.
[tex]\displaystyle\sum_{n\ge1}\frac1n[/tex]
Notice that this series essentially gives the left-endpoint Riemann sum approximation to the integral
[tex]\displaystyle\int_1^\infty\frac{\mathrm dx}x[/tex]
Because [tex]f(x)=\dfrac1x[/tex] is strictly decreasing for [tex]x>0[/tex], it follows that this approximation is greater than or equal to the value of the integral:
[tex]\displaystyle\sum_{n\ge1}\frac1n\ge\int_1^\infty\frac{\mathrm dx}x[/tex]
So if the integral diverges, then so must the finite series. You have
[tex]\displaystyle\int_1^\infty\frac{\mathrm dx}x=\lim_{t\to\infty}\int_1^t\frac{\mathrm dx}x=\ln|x|\bigg|_{x=1}^{x=t}[/tex]
[tex]=\displaystyle\lim_{t\to\infty}\ln|t|=\infty[/tex]
and thus the series must also diverge.
