Intuition for why or how can the series $1/n$ diverge, another perspective

Question

Please provide uncommon Answers

I was wondering how the series $H(n) = 1 + 1/2 + 1/3 +\dots+1/n$ diverges.
I suppose $\lim_{n\to \infty} (1/n)$ should be $0.$
Then, we find $H(n) - H(n-1) = 1/n.$ Taking the limit should render that,
$H(n) - H(n-1) \to 0$
Thereby we should be able to say, $H(n) = H(n-1).$ I know this is false, but how so?
Then how can it be divergent, as it after a point the graph is basically a straight line .
Yes I see the various proofs but fail to see an intuitive answer. Perhaps something fundamental is wrong with my thinking.
Please provide a complete explanation , Thank-you. As for intuition, I mean a way for me to easily see that it cannot diverge, without much manipulation of sorts, an unconventional way perhaps.

Answer 1

We consider the harmonic series: $$ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots $$

We can group the terms in a specific way:

$$ \begin{align*} 1 & \\ + \frac{1}{2} & \\ + \left( \frac{1}{3} + \frac{1}{4} \right) & \\ + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) & \\ + \left( \frac{1}{9} + \cdots + \frac{1}{16} \right) & \\ + \cdots \end{align*} $$

Now observe that in each group, each term is at least as large as the smallest term in the group. So we estimate:

• In the 3rd group: $$ \frac{1}{3} + \frac{1}{4} > 2 \cdot \frac{1}{4} = \frac{1}{2} $$
• In the 4th group: $$ \frac{1}{5} + \cdots + \frac{1}{8} > 4 \cdot \frac{1}{8} = \frac{1}{2} $$
• In the 5th group: $$ \frac{1}{9} + \cdots + \frac{1}{16} > 8 \cdot \frac{1}{16} = \frac{1}{2} $$

In general, for the $k$-th group (starting from the 3rd), there are $2^{k-2}$ terms, each at least $ \frac{1}{2^k} $, so the sum of the group is greater than: $$ 2^{k-2} \cdot \frac{1}{2^k} = \frac{1}{4} $$

(And with a looser bound, each group is greater than $ \frac{1}{2} $.)

Since there are infinitely many such groups, and each contributes a positive amount bounded below, the sum of the harmonic series diverges: $$ \sum_{n=1}^{\infty} \frac{1}{n} \to \infty $$

Answer 2

I'm not sure I really understand what you mean by "intuitive", but I can provide the simplest of proofs. But before doing so, if you consier the sequence $(u_n)_{n\in\mathbb{N}^*}=(\ln(n))_{n\in\mathbb{N}^*}$, $u_{n+1}-u_n\overset{n\to\infty}{\longrightarrow}0$ yet $(u_n)$ diverges, so your intuition here is wrong.

Note how, for all $x>-1$, $\ln(1+x)\leq x$. Now let $n\in\mathbb{N}$.

$$\ln(n+1)=\sum_{k=1}^n\ln(k+1)-\ln(k)=\sum_{k=1}^n\ln(1+1/k)\leq\sum_{k=1}^n\frac{1}{k}=H_n$$

Hence the divergence. I think this is the simplest proof one can find...

For an "intuitive" proof, maybe check Oresme's original method?

Answer 3

https://www.youtube.com/watch?v=aKl7Gwh297c

See this video from 5:18 mins onward. This provides a calculus based intuition.