Show that $\lim_{n}\sum_{k=n}^{2n}{1\over k} = \ln2$ using elementary methods. 
Prove that:
  $$
\lim_{n\to\infty}\left({1\over n} + {1\over n+1} + \cdots + {1\over 2n} \right) = \ln2
$$

I would like to show that using elementary methods, since I'm not allowed to even use derivatives, not to mention integrals.
Before this one, I've been able to show that:
$$
\exists\lim_{n\to\infty}\left(1 + {1\over 2} + {1\over 3} + \cdots + {1\over n} - \ln n\right) = L \tag 1
$$
Since the expression in $(1)$ under the sign of limit is bounded below and is monotonically decreasing, then by monotone convergence theorem it must converge to some number (which appeared to be named the Euler-Mascheroni constant.)
Now since $(1)$ converges then it must satisfy Cauchy's Criteria. Let's define the following sequence:
$$
x_n = 1 + {1\over 2} + {1\over 3} + \cdots + {1\over n} - \ln n
$$
Then $x_{2n}$ is defined as follows:
$$
x_{2n} = 1 + {1\over 2} + {1\over 3} + \cdots + {1\over 2n} - \ln (2n)
$$
But both limits exist and are equal, which implies that:
$$
\exists\lim_{n\to\infty}|x_{2n} - x_n| = 0 \tag2
$$
Now performing some algebraic manipulations on $(2)$ one may obtain:
$$
\begin{align}
|x_{2n} - x_n| &= \left|\sum_{k=1}^{2n}{1\over k} - \sum_{k=1}^{n}{1\over k} -\ln(2n) + \ln n\right|\\
&=\left|\sum_{k=n+1}^{2n}{1\over k} - (\ln(2n) -\ln n)\right| \\
&=\left|\sum_{k=n+1}^{2n}{1\over k} - \ln 2\right|
\end{align}
$$
Since $|x_{2n} - x_n|$ is convergent to $0$ then:
$$
\forall \epsilon > 0\ \exists N\in\Bbb N: \forall n\ge N \implies |x_{2n} - x_n| = \left|\sum_{k=n+1}^{2n}{1\over k} - \ln 2\right| < \epsilon
$$
Which is a standard definition of the limit, hence:
$$
\lim_{n\to\infty}\sum_{k=n}^{2n}{1\over k} = \ln 2
$$
I would like to ask for verification of the proof above, and/or point to mistakes in case of any. Thank you!
Note: This problem is given among other problems in the "Limit of numerical sequences" section. Long before the definition of the Integral is given.
 A: Your argument is ok if you take $(1) $ as the starting point. I would say that your writing is too convoluted, and you are misusing the $\exists $ symbol.
You could simply say
\begin{align}
\sum_{k=n+1}^{2n}\frac1k-\log2=\left (\sum_{k=1}^{2n}\frac1k-\log2n\right)-\left (\sum_{k=1}^{n}\frac1k-\log n\right)\xrightarrow [n\to\infty]{}L-L=0.
\end{align}
You need to distinguish between how you get the idea, and how you write the proof.
A: I do not know if this is an elementary method.
$$S_n=\sum_{k=0}^n \frac 1{n+k}=H_{2 n}-H_{n-1}$$ Now, let us use the asymptotics of harmonic numbers
$$H(p)=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ Apply it to each term and continue with Taylor expansion to get
$$S_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$
A: 
It can be found by considering the underlying Riemann Sums:

$\displaystyle\lim_{n \to \infty}\sum_{k = n}^{2n}{1 \over k} =
\lim_{n \to \infty}\left({1 \over n}\sum_{k = n}^{2n}{1 \over k/n}\right) = \int_{1}^{2}{\mathrm{d}k \over k} = \bbox[10px,border:1px groove navy]{\ln\left(2\right)}$
A: Some of the users ask how I proved $(1)$ without the definition of an Integral, which is too long for a comment. 
Some time ago I've shown that the following limit exists:
$$
\lim_{n\to\infty}\left(1 + {1\over n}\right)^n = e
$$
Later I've shown that:
$$
\lim_{n\to\infty}\left(1 + {k\over n}\right)^n = e^k
$$
I've also shown that $\left(1 + {1\over n}\right)^{n+1}$ is monotonically decreasing and tends to $e$, and $\ln(1+n) \le n$. I've used these facts to obtain a proof below.
By:
$$
\ln(1+n) \le n,\ n\in\Bbb N
$$
we have:
$$
1 + {1\over 2} + {1\over 3} + \cdots + {1\over n} - \ln n \ge \ln(1+1) + \ln\left(1+ {1\over 2}\right) + \cdots \ln\left(1+ {1\over n}\right) - \ln n
$$
Now by telescoping:
$$
\ln\left({2\over 1}\right) + \ln\left(3\over 2\right) + \cdots + \ln\left({n+1\over n}\right) - \ln n = \\
\ln\left(\frac{2\cdot 3\cdot 4\cdots\cdot(n+1)}{1\cdot 2\cdot 3\cdots n}\right) - \ln n =\\
\ln(n+1) - \ln n \ge 0
$$
So $x_n$ is bounded below. Then:
$$
\begin{align}
x_{n+1} - x_{n} &= {1\over n+1} - \ln(n+1) + \ln n \\
&={1\over n+1} - (\ln (n+1) - \ln n) \\
&= {1\over n+1} - \ln\left(1 + {1\over n}\right)
\end{align}
$$
We can now show that:
$$
\begin{align}
\left(1 + {1\over n}\right)^{n+1} &\ge e \iff \\
\ln\left(1 + {1\over n}\right)^{n+1} &\ge 1 \iff \\
\ln\left(1 + {1\over n}\right) &\ge {1\over n+1}
\end{align}
$$
So:
$$
{1\over n+1} - \ln\left(1 + {1\over n}\right) \le 0
$$
Which means $x_n$ is monotonically decreasing. Finally by monotone convergence theorem $x_n$ is convergent.
