alternative way of proving $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ converges to $\ln 2$ without using integrals I found a lot of answered to this problem using Riemann - sums, I myself solved it rewriting the logarithm sum, but there is another way where you should use the following hints.

Show that
$$\exp \left( \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n} \right) =
 2$$
using the following identities
(1). There is a nonnegative sequence $x_k$, converging to zero, and
$$\exp \left(\frac{1}{k} \right) = \left(1+\frac{1}{k} \right)\times
 \exp\left(\frac{x_k}{k} \right)$$
(2). $$\lim_{x \to 0} \frac{\exp(x)-1}{x}=1$$

To prove $(1)$, I wanted to rewrite
$$\exp\left(\frac{1}{k} \right)\times \frac{1}{1+\frac{1}{k}}$$ and expand $$\frac{1}{1+\frac{1}{k}} = \frac{1}{1-\frac{-1}{k}}$$  as geometric sequence to find $x_k$, but I don’t get anything useful.
For $(2)$ I thought may one should “stretch”
$$\sum_{k=1}^n \left(\exp\left(\frac{1}{k+n}\right)-1 \right)$$
which has the same limit (I proved that) as $\sum_{k=1}^n \frac{1}{k+n}$ and rewrite it as
$$\sum_{k=1}^n \frac{\exp\left(\frac{1}{k+n}\right)-1}{\frac{1}{k+n}}\frac{1}{k+n}$$
I’m thankful for any tips or answers!

Using the inequality$$\frac{1}{(n+1)}<\ln(\frac{n+1}{n})<\frac{1}{n}$$( Since $\frac{1}{(n+1)}*1<\ \int_{n}^{n+1}\frac{1}{x} \,dx <\frac{1}{n}*1$ (think area graphically) ),we can write $\ln(\frac{n+1}{n})<\frac{1}{n}$$<$$\frac{1}{n-1}<ln(\frac{n-1}{n-2})$. Then sum from $n$ to $2n$ and take $\lim_{n\to\infty}$.
$ \sum_{i=0}^{n}ln(\frac{n+i+1}{n+i})$$<$ $\sum_{i=0}^{n} \frac{1}{n+i} $$<$$ \sum_{i=0}^{n} ln(\frac{n+i-1}{n+i-2})$
=$\ln(\frac{n+1}{n}*\frac{n+2}{n+1}*...*\frac{2n+1}{2n})<\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n}<ln(\frac{n-1}{n-2}*\frac{n}{n-1}*...*\frac{2n-1}{2n-2})$
= $\ln(\frac{2n+1}{n})<\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n}<ln(\frac{2n-1}{n-2})$
$\lim_{n\to\infty}$$\ln(\frac{2n+1}{n})<$$\lim_{n\to\infty}$$(\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n})<$$\lim_{n\to\infty}$$ln(\frac{2n-1}{n-2})$
$\lim_{n\to\infty}$$\ln(2+\frac{1}{n})<$$\lim_{n\to\infty}$$(\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n})<$$\lim_{n\to\infty}$$ln(\frac{2-\frac{1}{n}}{1-\frac{2}{n}})$
You will see by the sandwich theorem the value comes to be $\ln2$.
