$s_n=\sum_{k=1}^n \frac{1}{k}$ then $s_n$ is bounded sequence?

Question

Let $s_n=\sum_{k=1}^n \frac{1}{k}$ then $s_n$ is bounded sequence?
I know that $\sum_{k=1}^\infty \frac{1}{k}$ is not convergent so sequence of its partial sum is divergent and $\sum_{k=1}^\infty \frac{1}{k}$=$\lim_{n\to\infty}s_n$

but i am not able to visualize this if i think in the following way
$\sum_{k=1}^\infty \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......$ = 1+0.5+0.33333..+0.25+..... in this sum first term is 1 and remaining all are 0.something then why it is not bounded ?

Answer 1

$\sum_{i=1}^{n}1/i $ is not bounded. (In fact the value of partial sum is strictly increasing, but grow extremely slow.)

Classic proof: (quite easy to see divergence by grouping terms)

$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+...$ $>1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+...=1+1/2+1/2+1/2+...$

Answer 2

$(s_n)$ is not bounded (another pro0f): suppose that $(s_n)$ is bounded an let $a:= \sup\{s_n: n \in \mathbb N\}$

From $\frac{1}{n+j} \ge \frac{1}{2n}$ we see that

$$s_{2n} \ge s_n+\frac{1}{2}$$

Thus $s_n+\frac{1}{2} \le a$ for all $n$, hence $s_n \le a-\frac{1}{2}$ for all $n$, a contradiction.

With $s_{2n} \ge s_n+\frac{1}{2}$ we can also prove that the harmonic series is divergent. Suppose that we have convegence an let $s$ be the limit of $(s_n)$

Then it follows $s \ge s+\frac{1}{2}$, which is absurd