Show that $\lim_{n}\sum_{k=n}^{2n}{1\over k} = \ln2$ using elementary methods.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $$S_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$

Answer 4

It can be found by considering the underlying Riemann Sums:

$\displaystyle\lim_{n \to \infty}\sum_{k = n}^{2n}{1 \over k} = \lim_{n \to \infty}\left({1 \over n}\sum_{k = n}^{2n}{1 \over k/n}\right) = \int_{1}^{2}{\mathrm{d}k \over k} = \bbox[10px,border:1px groove navy]{\ln\left(2\right)}$