Evaluating $\int_ 1^2\frac{1}{x}dx$ with a Riemann Sum I'm trying to solve
$$ \int_ 1^2 \frac{1}{x} \ dx $$
using Riemann sums, however I'm having trouble solving it.
 A: $$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}} = \sum_{k=1}^{n}\frac{1}{n+k}=H_{2n}-H_n = \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}\tag{1}$$
and:
$$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k}=\left.\log(1+x)\right|_{x=1^-}=\color{red}{\log 2}.\tag{2}$$
A: Hint. One may consider, for $n\ge 1$,
$$
\frac1n\sum_{k=0}^n \frac1{1+\frac{k}n}=\sum_{k=0}^n \frac1{n+k}=\sum_{k=n}^{2n} \frac1k=\ln 2+(H_{2n}-\ln (2n))-(H_n-\ln n) \tag1
$$ and one may use that
$$
\lim_{n \to \infty}\left(H_n-\ln n \right)=\gamma, \tag2
$$ the Euler-Mascheroni constant.
Then $(1)$ and $(2)$ give, as $n \to \infty$, 
$$
\frac1n\sum_{k=0}^n \frac1{1+\frac{k}n} \to  \ln 2 
$$ which is also $$\int_1^2 \frac1x\:dx.$$
A: $$
\begin{align}
\int_1^2\frac1x\,\mathrm{d}x
&=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac nk\cdot\frac1n\\
&=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac2{2k}\right)\\
&=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k\tag{1}
\end{align}
$$
By the alternating series test, the limit in $(1)$ exists. Call it $\alpha$.
In the derivation of $(1)$, we have the equation
$$
\begin{align}
\int_1^2\frac1x\,\mathrm{d}x
&=\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+k}\\
&=\alpha\tag{2}
\end{align}
$$

In this answer, It is shown that $\left(1+\frac1n\right)^n$ increases and $\left(1+\frac1n\right)^{n+1}$ decreases to $e$. Thus,
$$
\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}\le\dots\le\left(1+\frac1{n+k}\right)^{n+k}\tag{3}
$$
and
$$
\left(1+\frac1n\right)^{n+1}\ge\left(1+\frac1{n+1}\right)^{n+2}\ge\dots\ge\left(1+\frac1{n+k}\right)^{n+k+1}\tag{4}
$$
Putting $(3)$ and $(4)$ together yields, for $0\le k\le n$,
$$
\left(1+\frac1{n+k}\right)^{\frac{n}{n+1}}\le\left(1+\frac1{n+k}\right)^{\frac{n}{n+k}\frac{n+k+1}{n+1}}\le\left(1+\frac1n\right)^\frac{n}{n+k}\le\left(1+\frac1{n+k}\right)\tag{5}
$$
Therefore,
$$
\begin{align}
e^\alpha
&=\lim_{n\to\infty}\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{6}
\end{align}
$$
and by $(5)$
$$
\begin{align}
\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
&\le\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\dots\left(1+\frac1{2n}\right)\\
&=\frac{2n+1}{n+1}\tag{7}
\end{align}
$$
Using the other direction of $(5)$, we get
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{8}
$$
By the Squeeze Theorem, $(6)$, $(7)$, and $(8)$ say
$$
e^\alpha=2\tag{9}
$$
Therefore,
$$
\int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{10}
$$
A: Choose the partition $x_k:=2^{k/N}$, $\>0\leq k\leq N$. Then
$$\eqalign{\int_1^2{1\over x}\>dx&\doteq\sum_{k=0}^{N-1}{1\over x_k}(x_{k+1}-x_k)=\sum_{k=0}^{N-1}\left({x_{k+1}\over x_k}-1\right)\cr&=N(2^{1/N}-1)={2^{1/N}-1\over 1/N}\to\log2\qquad(N\to\infty)\ .\cr}$$
A: Let $P=\{x_0, x_1,\cdots,x_n\}$ be any partition of $[1,2]$, 
and let $\displaystyle c_i=\frac{x_i-x_{i-1}}{\ln(x_i)-\ln(x_{i-1})}$ for $1\le i\le n;\;\;$ so $c_i\in[x_{i-1},x_i]$ for each $i$.
Then $\displaystyle\sum_{i=1}^n f(c_i)\Delta x_i=\sum_{i=1}^n\frac{\ln(x_i)-\ln(x_{i-1})}{x_i-x_{i-1}}(x_i-x_{i-1})=\sum_{i=1}^n(\ln(x_i)-\ln(x_{i-1})=\ln2-\ln 1=\color{red}{\ln 2}$
