How do I prove that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges?

Question

I am studying infinite series in class. Our teacher showed us how the series $$\sum_{n=1}^\infty \frac{1}{n}=1+1/2+1/3+...$$ diverges because $$\int_{1}^\infty \frac {1}{n}dn=\infty$$ Is there a better way to show how the above series diverges?

You can also prove (by induction) that $$ s_{2^k} \geq 1+ k/2 $$ where $s_n$ is the sequence of partial sums. — Prahlad Vaidyanathan, Oct 24 '13 at 02:33
A nice way of seeing it is by the comparison test found here http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29 — JessicaK, Oct 24 '13 at 02:35
There are several ways to show that the harmonic series divreges; look at this amusingly titled article: scipp.ucsc.edu/~haber/archives/physics116A10/harmapa.pdf‎ — knsam, Oct 24 '13 at 02:36
For more fun, let $x$ be the sum. Note that $1+\frac{1}{3}+\frac{1}{5}+\cdots\gt \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots=\frac{x}{2}$. Thus $x\lt 2\cdot\frac{x}{2}$. — André Nicolas, Oct 24 '13 at 02:49
Note that $\log_2 n \le \lceil \log_2 n \rceil$, and so $n = 2^{\log_2 n} \le 2^{\lceil \log_2 n \rceil}$.
Then $\sum_n \frac{1}{n} \ge \sum_n \frac{1}{2^{\lceil \log_2 n \rceil}} = 1+\sum_{k=0}^\infty \sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2^{\lceil \log_2 i \rceil}} \ge 1+\sum_{k=0}^\infty \sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} = 1 + \sum_{k=0}^\infty \frac{1}{2}$. — copper.hat, Oct 24 '13 at 02:52

score 5 · Answer 1 · answered Oct 24 '13 at 02:36

"Better" way? What does "better" mean? In a sense, that explanation is quick and simple. How do you beat that?

There is a cute way. Note that $1/3$ and $1/4$ are both $\geq 1/4$. So that $1/3 + 1/4 \geq 1/2$. Similarly, $1/5, 1/6, 1/7, 1/8 \geq 1/8$, so $\frac15 + \frac16 + \frac17 + \frac18 \geq \frac12$. The next $8$ terms will also be bigger than $\frac12$. So will the next $16$, and so on.

In this way, we see that $\displaystyle \sum \frac1n > \sum \frac12$, which clearly diverges.

Is this better? I don't think so. But I think it's cute.

Perhaps a better way would be to better examine the integral one. What the integral really shows is that $$ \sum_{n = 1}^x \frac{1}{n} \approx \int_1^x \frac{1}{n}\mathrm{d}n = \log x.$$

So not only do we know that it gets very big, but we know exactly how quickly it diverges. That's something special, arguably better.

score 1 · Answer 2 · answered Oct 24 '13 at 02:40

Suppose that it were convergent, and define $$f_n = n (n + 1) \chi_{[\frac{1}{n + 1}, \frac{1}{n}]}$$ Then $$\int_0^1 f_n dm = 1$$

for each $n$, but $f_n \to 0$ pointwise everywhere on $[0, 1]$. Define $$g = \sum_n f_n$$

Since we've assumed the series to be convergent, $g$ is integrable. Hence by the Dominated Convergence Theorem,

$$0 = \int_0^1 \lim_n f_n dm = \lim_n \int_0^1 f_n dm = \lim_n 1 = 1$$

1233dfv · Answer 3 · 2013-10-24T19:26:29.527

We will consider the partial sums $s_1$, $s_2$, $s_4$, $s_8$, $s_{16}$,... of the Harmonic Series and discover something very interesting about the way it behaves.

$s_1=1$

$s_2=1+{1\over2}$

$s_4=1+{1\over2}+{1\over3}+{1\over4}>1+{1\over2}+({1\over4}+{1\over4})=1+{2\over2}$

$s_8=1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+{1\over7}+{1\over8}>1+{1\over2}+({1\over4}+{1\over4})+({1\over8}+{1\over8}+{1\over8}+{1\over8})=1+{3\over2}$

$s_{16}=1+{1\over2}+{1\over3}+{1\over4}+\cdots+{1\over15}+{1\over16}>1+{1\over2}+({1\over4}+{1\over4})+({1\over8}+\cdots+{1\over8})+({1\over16}+\cdots+{1\over16})=1+{4\over2}$

We can see that $s_{2n}>1+{n\over2}$ from the above argument. Thus as $n\rightarrow\infty$ we can see that $s_{2n}\rightarrow\infty$ and we can conclude that the Harmonic Series $\sum_{n=1}^\infty \frac{1}{n}$ is divergent.

How do I prove that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges?

3 Answers3