I have to show without using any derivative that:
$$i) \; \lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{1}{n+i-1}=\int_1^2\frac{1}{x}\,dx$$
And, using the Fundamental Theorem of Calculus:
$$ii) \;\text{If} \; r\in \mathbb{N}, \;\text{then}\; \lim_{n\rightarrow \infty}\sum_{i=1}^n\frac{i^r}{n^{r+1}}=\frac{1}{r+1}.$$
So, from
$f\in R([a,b])\leftrightarrow$ $f$ is integrable in the sense of Riemann's definition.
We have the next Corollary:
$$\int_a^bf\,d\alpha=\lim_{n\rightarrow\infty}\sum_{i=1}^{N(n)}f\left(\xi_i^n\right)\cdot \left(\alpha( x_i^n)- \alpha( x_{i-1}^n)\right)$$
So, I'm trying to follow an example done in class but don't quite get it:
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n}=\frac{1}{2n}\left( \frac{1}{\frac{1}{2}+\frac{1}{2n}}+\frac{1}{\frac{1}{2}+\frac{2}{2n}}+\cdots+\frac{1}{\frac{1}{2}+\frac{2n}{2n}}\right)$$
And since $f(x)=\frac{1}{\frac{1}{2}+x}$ is continuous on $[0,1]$, we get $f\in R\left(\left[0,1 \right]\right)$
Then:
$$\lim_{n\rightarrow \infty} \left( \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n} \right)=\lim_{n\rightarrow \infty}\sum_{k=1}^{2n}f\left( \frac{k}{2n} \right) \left( \frac{k}{2n}-\frac{k-1}{2n} \right)=\int_0^1\frac{1}{\frac{1}{2}+x}\,dx.$$
What kind of $\alpha(x)$ would I need for my particular problem $(i)$? What does it has to satisfy in order for the last equality to hold?
I guess that understanding $i)$ will make $ii)$ easier, but.. is there something trivial that I'm not seeing?
(1) Let $f:\Bbb R_{>0}\to\Bbb R$ be the function $f(x)=1/x$. Then $$ \begin{aligned} \frac 1{n+0}+ \frac 1{n+1}+\dots+ \frac 1{n+(n-1)} &= \frac 1n\left( \underbrace{\frac 1{1+ \frac 0n}}_{f\left(1+\frac 0n\right)} +\underbrace{\frac 1{1+ \frac 1n}}_{f\left(1+\frac 1n\right)} +\underbrace{\frac 1{1+ \frac {n-1}n}}_{f\left(1+\frac {n-1}n\right)} \ \right) &\to\int_1^2 f(x)\; dx\ . \end{aligned} $$ Above, $1+\frac 0n$, $1+\frac 1n$, $\dots$, $1+\frac {n-1}n$, are intermediate points in the interval $[1,2]$ in the obvious equidistant intervals of length $\frac 1n$, so the Riemann sums above converge to the Riemann integral of $f$ on $[1,2]$.
(2) Use the Rieman sums convergence to the Riemann integral for the function $x\to x^r$, considered on $[0,1]$, so rewrite $$ \sum_{1\le k\le n} \frac {k^r}{n^{r+1}} = \frac 1n\sum_{1\le k\le n} \left(\frac kn\right)^r \to \int_0^1x^r\;dx =\frac 1{r+1}\ . $$