Prove that $\lim_{n\to\infty} H_n/n = 0$ ($H_n$ is the $n$-th harmonic number) using certain techniques I can't seem to use certain methods such as $\varepsilon$-N, L'Hôspital's Rule, Riemann Sums, Integral Test and Divergence Test Contrapositive or Euler's Integral Representation to prove that
$\lim_{n-> \infty} \frac{H_n}{n} = 0$ where $H_n$ is the nth Harmonic number $= \sum_{i=1}^{n} \frac{1}{i} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n}$?
I was able to prove it using the Monotone Convergence Theorem, and I think polylogarithms make this easy. I would like to know if any of these are correct/can be modified to be correct/even possible to use.
Here are my attempts:
1 $\varepsilon-\delta$
$\forall \varepsilon > 0, \exists N > 0$ s.t. $|\frac{H_n}{n} - 0| < \varepsilon$ whenever $n > N$
$|\frac{H_n}{n} - 0|$
$=|\frac{H_n}{n}|$
$ \leq |\frac{H_n}{n}|$
$ \leq \frac{|H_n|}{|n|}$
$ \leq \frac{|H_n|}{|n|} < \varepsilon$ if $n > N=\frac{H_n}{\epsilon}$
But I haven't really isolated n, so I don't think that's allowed?
2 L'Hôspital's Rule
$\lim_{n-> \infty} \frac{H_n}{n}$
$= \lim_{n-> \infty} \frac{\frac{\partial}{\partial n} H_n}{1}$
Is there a discrete version of the Fundamental Theorem of Calculus that allows us to evaluate $\frac{\partial}{\partial n} H_n$? Maybe it's this one, but I don't understand it. Does $\frac{\partial}{\partial n} H_n$ even have meaning?
Or is there some way to show that $\lim_{n-> \infty} \frac{H_n}{n} = \lim_{n-> \infty} \frac{1}{n}\int_{1}^{n} \frac{1}{x} dx$?
3 Riemann Sums
$\lim_{n-> \infty} \frac{H_n}{n}$
$= \lim_{n-> \infty} \frac{\sum_{i=1}^{n}\frac{1}{i}}{n}$
$= \lim_{n-> \infty} \frac{1}{n}\sum_{i=1}^{n}\frac{1}{i}$
$= \lim_{n-> \infty} \frac{1}{n}\sum_{i=1}^{n}\frac{1}{i/n}\frac{1}{n}$
$= \lim_{n-> \infty} \frac{1}{n} \lim_{n-> \infty} \sum_{i=1}^{n}\frac{1}{i/n}\frac{1}{n}$
$= \lim_{n-> \infty} \frac{1}{n} \int_{0}^{1} \frac{1}{x} dx$
$= \lim_{n-> \infty} \frac{1}{n} ln|x||_{0}^{1}$
$= \lim_{n-> \infty} \frac{1}{n} (ln|1| - ln|0|)$
$= \lim_{n-> \infty} \frac{1}{n} (- ln|0|)$
$= \lim_{n-> \infty} \frac{1}{n} (- ln|1/n|)$ if that's even allowed
$= \lim_{n-> \infty} \frac{1}{1} (- 1/(1/n)) \lim_{n-> \infty} \frac{-1}{n^{2}}$ by L'Hôspital's Rule
$= \lim_{n-> \infty} \frac{1}{n^{2}} (1/(1/n))$
$= \lim_{n-> \infty} \frac{1}{n^{2}} (n)$
$= \lim_{n-> \infty} \frac{1}{n} = 0$
4 Integral Test and Divergence Test Contrapositive
If the series $\sum_{n=1}^{\infty} \frac{H_n}{n}$ is convergent then $\lim_{n-> \infty} \frac{H_n}{n} = 0$
If the integral $\int_{1}^{\infty} \frac{H_x}{x} dx$ is convergent then the series $\sum_{n=1}^{\infty} \frac{H_n}{n}$ is convergent.
I don't know how to integrate $\int_{1}^{\infty} \frac{H_x}{x} dx$.
How about $\int_{1}^{\infty} \frac{\int_{1}^{x} \frac{1}{y} dy}{x} dx$ ? If that is convergent does that mean the series $\sum_{n=1}^{\infty} \frac{H_n}{n}$ is convergent? It doesn't seem to be convergent in the first place though:
$\int_{1}^{\infty} \frac{\int_{1}^{x} \frac{1}{y} dy}{x} dx$
$= \int_{1}^{\infty} \int_{1}^{x} \frac{1}{xy} dy dx$
$= \int_{1}^{\infty} \frac{ln|y|}{x}|_{y=1}^{y=x} dx$
$= \int_{1}^{\infty} \frac{ln|x|}{x} dx$
$= \int_{1}^{\infty} \frac{lnx}{x} dx$
$= (lnx)^{2}|_{1}^{\infty} = \infty$
5 Euler's Integral Representation
Euler proved that $H_n = \int_{0}^{1} \frac{1-x^{n}}{1-x} dx$
So, $\lim_{n-> \infty} \frac{H_n}{n}$
$= \lim_{n-> \infty} \frac{\int_{0}^{1} \frac{1-x^{n}}{1-x} dx}{n}$
$= \lim_{n-> \infty} \frac{\partial}{\partial n}\int_{0}^{1} \frac{1-x^{n}}{1-x} dx$ by L'Hôspital's Rule
$= \lim_{n-> \infty} \int_{0}^{1} \frac{\partial}{\partial n} \frac{1-x^{n}}{1-x} dx$
$= \lim_{n-> \infty} \int_{0}^{1} \frac{\partial}{\partial n} \frac{1-x^{n}}{1-x} dx$
$= \lim_{n-> \infty} \int_{0}^{1} \frac{\partial}{\partial n} \frac{-x^{n}}{1-x} dx$
$= \lim_{n-> \infty} \int_{0}^{1} \frac{-nx^{n-1}}{1-x} dx$
$= \lim_{n-> \infty} \int_{0}^{1} \frac{nx^{n-1}}{x-1} dx$
$= \lim_{n-> \infty} n \int_{0}^{1} \frac{x^{n-1}}{x-1} dx$
$= \lim_{n-> \infty} n \int_{0}^{1} \frac{x^{n-1} - 1}{x-1} + \frac{1}{x-1}dx$
$= \lim_{n-> \infty} n \int_{0}^{1} \sum_{i=0}^{n-2} x^{i}+ \frac{1}{x-1}dx$
It looks like I will get an ln(0) again...
 A: This is almost obvious if one remembers the following elementary (but not so easy to prove) result:

If $a_{n} \to L$ as $n \to \infty$ then $\dfrac{a_{1} + a_{2} + \cdots + a_{n}}{n} \to L$ as $n \to \infty$.

Clearly here $a_{n} = 1/n \to 0$ as $n \to \infty$ and $H_{n} = a_{1} + a_{2} + \cdots + a_{n}$ so that $H_{n}/n \to 0$ as $n \to \infty$.
A: We have $$\ln(n+1)\le H_n\le\ln(n)+1\tag{1}$$
$$\frac{\ln(n+1)}n\le \frac{H_n}{n}\le\frac{\ln(n)+1}{n}$$
Both the right and left sides tend to zero as $n\to\infty$ using L'hospital. And therefore by the squeeze theorem, our limit is $0$

$$\int_1^{n+1} f(x)\,{\rm d}x \le \sum_{i=1}^nf(i)\le 1+\int_1^n f(x)\,{\rm d}x\text{ for monotonically decreasing $f$}$$
Here $f(x)=\frac{1}{x}$
A: Although this is an old answered post, an unaddressed question is whether we can formulate a proof using the Cauchy definition of limit.  The answer is yes.  The proof relies on the observation that for every positive integer $n$, $$\sum_{k=2^{n-1}+1}^{2^n} \frac{1}{k} < \sum_{k=2^{n-1}+1}^{2^n} \frac{1}{2^{n-1}} = 1$$ which easily follows because $k > 2^{n-1}$ for every $k \in \{2^{n-1} + 1, \ldots, 2^n\}$.  Consequently, $$H_{2^n} = 1 + \sum_{m=1}^n \sum_{k=2^{m-1} +1}^{2^m} \frac{1}{k} < 1 + \sum_{m=1}^n 1 = n+1.$$  Therefore, $$0 < \frac{H_{2^n}}{2^n} < \frac{n+1}{2^n} $$  This suggests that for sufficiently small $\epsilon$, the choice $N = 2^{1/\epsilon}$ will give the desired bound (the details of establishing the resulting inequality is left to the reader as an exercise).
A: The discrete version of L'Hôpital is Stolz:
$$\lim_{n\to\infty}\frac{H_n}{n} =
\lim_{n\to\infty}\frac{H_{n+1}-H_n}{(n+1)-n} =
\lim_{n\to\infty}\frac{\frac{1}{n+1}}{1} = 0.
$$
