0

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 ?

Martin Sleziak
  • 53,687
dipali mali
  • 555
  • Hard to believe? Well, some things, hard to believe, are true. Going by your logic, you could ask : Why isn't $\sum_{i=1}^n \frac {0.00000000000000000000000000001}n$ bounded? Isn't each term very small? The harmonic series isn't decreasing fast enough, although it's terms are small. The right way to visualize this, is to look at the way this function exceeds the natural logarithm function (or a scalar multiple), which you can plot on Wolfram Alpha and see for yourself. – Sarvesh Ravichandran Iyer Nov 10 '16 at 06:22

2 Answers2

3

$\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)

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

Nick
  • 2,769
0

$(s_n)$ 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

Fred
  • 77,394