For $n\geq 2$, let $x_n$ denote the smallest integer such that $\displaystyle \sum_{k=n}^{x_n} \frac 1k >1$. Prove that $\lim_n \frac{x_n}{n}\sim e$.
I've found this problem in an old handout of mine. $x_n$ is well-defined since the harmonic series diverges. By definition of $x_n$, $\sum_{k=n}^{x_n-1} \frac 1k\leq 1$ and $\sum_{k=n}^{x_n} \frac 1k >1$, thus $$1+\frac{1}{x_n}\geq \sum_{k=n}^{x_n} \frac 1k >1$$
That's all I've found so far.
Combine your inequalities with integral inequalities to get
$$\int_n^{x_n-1}\frac1t~{\rm d}t<\sum_{k=n}^{x_n-1}\frac1k<1<\sum_{k=n}^{x_n}\frac1k<\frac1n+\int_n^{x_n}\frac1t~{\rm d}t$$
Evaluating gives us
$$\ln\left(\frac{x_n}n-\frac1n\right)<1<\frac1n+\ln\left(\frac{x_n}n\right)$$
Solving for $x_n/n$ gives us
$$\exp\left(1-\frac1n\right)<\frac{x_n}n<e+\frac1n$$
And as $n\to\infty$, we get the squeeze desired.