# $x_n$ is the smallest number for which $\sum_{i=1}^{x_n}\frac{1}{i}\geq n$ .Find $\lim_{n \to \infty}\frac{x_{n+1}}{x_n}$.

$$(x_n)_{n\gt1}$$ is a sequence such that $$x_n$$ is the smallest number for which $$\sum_{i=1}^{x_n}\frac{1}{i}\geq n$$ is true.
Find $$\lim_{n \to \infty}\frac{x_{n+1}}{x_n}$$.
I looked on OEIS - A002387 for the values in the sequence and figured out the limit should be $$e$$.

What I tried:
From $$H_n=\sum_{i=1}^{n}\frac{1}{i}=\ln n\ + c_n$$ I obtained $$n = e^{{H_n}-c_n}$$ so $$x_n = e^{{H_{x_n}}-c_{x_n}}$$ and from here
$$\frac{x_{n+1}}{x_n}=e^{{H_{x_{n+1}}}-c_{x_{n+1}}-{H_{x_n}}+c_{x_n}}$$.
As $$n \to \infty$$, $$c_n\to\gamma$$ (and because $$x_n\to\infty$$ as $$n \to \infty$$) $$\implies c_{x_n} \to\gamma$$.
So if I could prove that $${H_{x_{n+1}}}-{H_{x_n}}\to1$$ as $$n\to\infty$$ the limit would be $$e^1$$.
I tried to find a lower and upper bound to use the squeeze theorem, but I only managed to show that $$2\geq{H_{x_{n+1}}}-{H_{x_n}}\geq0$$ by using $$n+1\gt\sum_{i=1}^{x_n}\frac{1}{i}\geq n$$.

How can I prove that $${H_{x_{n+1}}}-{H_{x_n}}\to 1$$?

By definition $$H_{x_n}$$ is the smallest harmonic number greater than $$n$$, and because $$x_n\geq n$$ it follows that $$n These bounds allow you to squeeze the desired limit for $$H_{x_{n+1}}-H_{x_n}$$.