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$(x_n)_{n\gt1}$ 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 ${H_{x_{n+1}}}-{H_{x_n}}\to 1$?

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

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