Evaluating $\int_ 1^2\frac{1}{x}dx$ with a Riemann Sum

Question

I'm trying to solve $$ \int_ 1^2 \frac{1}{x} \ dx $$ using Riemann sums, however I'm having trouble solving it.

Answer 1

$$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}} = \sum_{k=1}^{n}\frac{1}{n+k}=H_{2n}-H_n = \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}\tag{1}$$ and: $$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k}=\left.\log(1+x)\right|_{x=1^-}=\color{red}{\log 2}.\tag{2}$$

Answer 2

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 $$\int_1^2 \frac1x\:dx.$$

Answer 3

$$ \begin{align} \int_1^2\frac1x\,\mathrm{d}x &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac nk\cdot\frac1n\\ &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac2{2k}\right)\\ &=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k\tag{1} \end{align} $$ By the alternating series test, the limit in $(1)$ exists. Call it $\alpha$.

In the derivation of $(1)$, we have the equation $$ \begin{align} \int_1^2\frac1x\,\mathrm{d}x &=\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+k}\\ &=\alpha\tag{2} \end{align} $$

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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, $$ \int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{10} $$

Answer 4

Choose the partition $x_k:=2^{k/N}$, $\>0\leq k\leq N$. Then $$\eqalign{\int_1^2{1\over x}\>dx&\doteq\sum_{k=0}^{N-1}{1\over x_k}(x_{k+1}-x_k)=\sum_{k=0}^{N-1}\left({x_{k+1}\over x_k}-1\right)\cr&=N(2^{1/N}-1)={2^{1/N}-1\over 1/N}\to\log2\qquad(N\to\infty)\ .\cr}$$

Answer 5

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 $\displaystyle\sum_{i=1}^n f(c_i)\Delta x_i=\sum_{i=1}^n\frac{\ln(x_i)-\ln(x_{i-1})}{x_i-x_{i-1}}(x_i-x_{i-1})=\sum_{i=1}^n(\ln(x_i)-\ln(x_{i-1})=\ln2-\ln 1=\color{red}{\ln 2}$