I have been tackling this for days and I am utterly confused on how to do this. I'd love to see how to work this out:
Let $0<a<b$ and $p$ be an element of the ℝ \ {-1}
a) Given a partition P = {$a=x_0< ... <x_n=b$} where $x_j=ar^{j/n}$ with $r=b/a$, show that for $f(x)=x^p$,
$\frac{b^{p+1}-a^{p+1}}{p+1}$ is equal to $\lim_{n\to∞}U(f,P_n)$ if $p>=0$, otherwise $\lim_{n\to∞}L(f,P_n)$ if $p<0$.
[Hint: Use geometric sums. You may have to compute $\lim_{t\to1}\frac{t-1}{t^{p+1}-1}$]
alet $P={t_{0},t_{1}, \cdots,t_{n} }$ a partition of $[a,b]$:
let's define $M_{i}=f(t_{i})$ with $f(x)=x^{p}$ and $t_{i}=ac^{i/n}$, then;
$$U(f,P)=\sum_{i=1}^{n}M_{i}(t_{i}-t_{i-1})=\sum_{i=1}^{n}(ac^{\frac{i}{n}})^{p}(ac^{\frac{i}{n}}-ac^{\frac{i-1}{n}})$$ $$=a^{p+1}(1-c^{-\frac{1}{n}})\sum_{i=1}^{n}(c^{\frac{p+1}{n}})^{i}$$ $$=a^{p+1}(1-c^{-\frac{1}{n}})\sum_{j=0}^{n-1}(c^{\frac{p+1}{n}})^{j+1}$$ $$=a^{p+1}(1-c^{-\frac{1}{n}})c^{\frac{p+1}{n}}\sum_{j=0}^{n-1}(c^{\frac{p+1}{n}})^{j}$$
then simplifying and using the geometric sum you have $$(a^{p+1}-b^{p+1})c^{\frac{p}{n}}(\frac{c^\frac{1}{n}-1)}{1-c^{(p+1)/n}})$$, the you have that: $$1+c^{1/n}+\cdots+c^p/n=\frac{1-c^{(n+1)/n}}{1-c^{1/n}}$$ $$\frac{-1}{1+c^{1/n}+\cdots+c^p/n}=\frac{c^{1/n}-1}{1-c^{(p+1)/n}}$$ $$=\frac{(b^{p+1}-a^{p+1})c^{p/n}}{1+c^{1/n}+\cdots+c^p/n}$$
similarly defines $L(f,P)$
to show that it is integrable, use the fact that; $$U(f,P)-L(f,P)<\epsilon$$, it's easy to see that $\int_{a}^{b}x^{p}=\frac{b^{(p+1)/n}-a^{(p+1)/n}}{1+p}$