TITLE: How to find the limit of a sum of reciprocals $\lim_{n\to\infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n})$?
QUESTION [6 upvotes]: There's a limit that I am unable to solve. I think it should be equal to $\infty$.
$$\lim_{n\to\infty}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$

REPLY [4 votes]: If $\sum \dfrac{1}{n}$ were convergent to $S$, then it is absolutely convergent.
As a result we have that
$S_1 = \sum \dfrac{1}{2n-1} = 1 + \dfrac{1}{3} +  \dfrac{1}{5} + \dots $
and
$S_2 = \sum \dfrac{1}{2n} = \dfrac{1}{2} + \dfrac{1}{4} + \dots$
are both absolutely convergent and
we have
$S_1 + S_2 = S$
and
$S_2 = S/2$ and thus
$S_1 = S_2$
which is not possible as $S_1 \gt S_2$.