How to find the limit of a sum of reciprocals $\lim_{n\to\infty}(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n})$? There's a limit that I am unable to solve. I think it should be equal to $\infty$.
$$\lim_{n\to\infty}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$
 A: Let $S_n = \sum_{k=1}^n \frac{1}{k}$, then $S_{2^{n+1}} - S_{2^n+1} \geq 2^n\frac{1}{2^{n+1}} = \frac{1}{2}$.
A: 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 $S_1 \gt S_2$.
A: Write $$s_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}$$
and observe that
$$s_{2n} - s_{n} = \frac{1}{n+1} + \frac{1}{n+2} + \cdots+ \frac{1}{2n} \geq \underbrace{\frac{1}{2n} + \cdots + \frac{1}{2n}}_{n\text{ terms}} = n \frac{1}{2n} = \frac{1}{2}$$
so that the sequence $(s_{n})_{n \in \mathbb{N}}$ can't be a Cauchy sequence, hence it can't converge. Put differently, $s_{2^{n}} \geq \frac{n}{2}$.

Edit:
What I was getting at seems not what Paul wanted to hear, but let me finish that argument nevertheless. It should be clear that $s_{n+1} = s_{n} + \frac{1}{n+1}$ so that $s_{n+1} \gt s_{n}$ or in words, the sequence $s_{n}$ is (strictly) monotonically increasing. I've argued that $s_{2^{n}} \geq \frac{n}{2}$ (see also Alexander's answer). Therefore we have $s_{k} \geq \frac{n}{2}$ for each $k \geq 2^{n}$ and as $\frac{n}{2}$ tends to infinity with $n$ so must $s_{k}$, as $k \to \infty$.
To find a divergent lower integral bound, note that the inequality $s_{n} \geq \int_{1}^{n} \frac{1}{x}\,dx = \log{n}$ follows from considering the upper Riemann sum associated to the partition $(1,2,\ldots,n)$ of $[1,n]$. Again this shows that $s_{n} \to \infty$ because $\log{n} \to \infty$ with $n \to \infty$ as Jonas explained in a comment to your question.
I hope that helps.
