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.
Best Answer
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.