Question

How to evaluate infinite series 1/n

Answer 1

This series diverges.

$\displaystyle\sum_{n\ge1}\frac1n$

Notice that this series essentially gives the left-endpoint Riemann sum approximation to the integral

$\displaystyle\int_1^\infty\frac{\mathrm dx}x$

Because $f(x)=\dfrac1x$ is strictly decreasing for $x>0$, it follows that this approximation is greater than or equal to the value of the integral:

$\displaystyle\sum_{n\ge1}\frac1n\ge\int_1^\infty\frac{\mathrm dx}x$

So if the integral diverges, then so must the finite series. You have

$\displaystyle\int_1^\infty\frac{\mathrm dx}x=\lim_{t\to\infty}\int_1^t\frac{\mathrm dx}x=\ln|x|\bigg|_{x=1}^{x=t}$
$=\displaystyle\lim_{t\to\infty}\ln|t|=\infty$

and thus the series must also diverge.