$1+\frac{1}{2}+...+\frac{1}{x_n}\geq n$. Determine $\lim\limits_{n\to\infty} \frac{x_{n+1}}{x_n}$ For every $n \in \mathbb{N}$ let $x_n$ be the smallest natural number such that:
$$1+\frac{1}{2}+...+\frac{1}{x_n}\geq n$$
Determine $\lim\limits_{n\to\infty} \frac{x_{n+1}}{x_n}$
It's the harmonic series but I can't figure when does the harmonic sum overcomes a number. I've done a script that finds the first 5 numbers from the sequence but I can't find a formula between them.
$$x_1=1\\x_2=4\\x_3=11\\x_4=31\\x_5=81$$
I believe it has to do something with exponential but I am not sure (from it's growth) 
 A: We will use the fact that for large $x_k$
$$\lim_{k\to\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x_k}- \log(x_k)\right)=\gamma$$
where $\gamma=0.577216$. Now, since 
$$\lim_{k\to\infty}\frac{1}{x_k}=0$$
we know that
$$\lim_{k\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x_k}-k)=0$$

EDIT: More details were requested for this step.
First, define
$$H_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$
Then we know that
$$H_{x_k}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x_k}\geq k$$
and
$$H_{x_k-1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x_k-1}\leq k$$
This implies
$$H_{x_k-1}\leq k\leq H_{x_k}$$
Thus
$$ 0=H_{x_k}-H_{x_k}\leq H_{x_k}-k\leq H_{x_k}-H_{x_k-1}=\frac{1}{x_k}$$
Since $x_k\geq k$ (and therefore goes to infinity) we know
$$0\leq\lim_{k\to\infty}(H_{x_k}-k)\leq \lim_{k\to\infty}\frac{1}{x_k}\leq \lim_{k\to\infty}\frac{1}{k}=0$$
We conclude 
$$\lim_{k\to\infty}(H_{x_k}-k)=0$$

This implies
$$\lim_{k\to\infty} \left(\log(x_{k+1})-\log(x_k)\right)$$
$$=\lim_{k\to\infty} \left(\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x_{k+1}}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x_k}\right)\right)$$
$$=\lim_{k\to\infty}\left((k+1)-k\right)=1$$
Thus
$$1=\lim_{k\to\infty} \left(\log(x_{k+1})-\log(x_k)\right)=\lim_{k\to\infty} \log\left(\frac{x_{k+1}}{x_k}\right)$$
Since the natural log is continuous at $1$, we may conclude
$$\lim_{k\to\infty}\frac{x_{k+1}}{x_k}=e$$
A: Here's a slightly different approach to the first answer that provides a more precise asymptotic:
I'll write $H_n = 1 + 1/2+ \ldots + 1/n$. It is known that
$$H_n = \log n + \gamma + o(1)$$
where $\gamma$ is the Euler-Mascheroni constant. There therefore exists two sequences $\delta_n, \epsilon_n =o(1)$ such that
$$\log n + \gamma + \delta_n \le H_n \le \log n + \gamma + \epsilon_n.$$
Since $H_{x_n} \ge n$,
$$\log x_n + \gamma + \epsilon_{x_n} \ge n$$
which rearranges to $x_n \ge e^{-\gamma} e^{n +o(1)}$. Further, since $H_{x_n} \ge n$ then we necessarily have that
$$\log x_n + \gamma + \delta_{x_n} < n + 1/x_n,$$
since, if this does not hold, $H_{x_n} \ge n+1/x_n$ and $H_{x_n -1} \ge n + 1/x_n - 1/x_n =n$ contradicting the definition of $x_n$ (since $x_n -1$ would be smaller than $x_n$ and still satisfy your inequality). Hence, $x_n < e^{-\gamma} e^{n+1/x_n +o(1)}$.
Combining our inequalities gives that
$$ \frac{x_n}{e^n} \sim e^{-\gamma}.$$
Your question follows easily from this asymptotic expression.
