How do I prove that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges? 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?
 A: "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.
A: 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,
$$0 = \int_0^1 \lim_n f_n dm = \lim_n \int_0^1 f_n dm = \lim_n 1 = 1$$
A: 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.
