Show that for any $\epsilon$ there exists $n$ so that $U(f,P_n)-L(f,P_n)<\epsilon$

Question

I have the upper and lower Riemann sums of the function $x^2$ on the interval $[1,b]$ with the partition $P_n: x_0=1, x_1=b^{1/n},...x_n=b^{n/n}=b$ for every positive integer $n$. The sums are equal to

$U(f,P_n)=b^{(2/n)} (b^{(1/n)}-1) \frac{(b^3-1)}{(b^{(3/n)}-1)}$ and $L(f,P_n)=(b^{(1/n)}-1) \frac{(b^3-1)}{(b^{(3/n)}-1)}$

so that

$U(f,P_n)-L(f,P_n)=(b^{(2/n)}-1) (b^{(1/n)}-1) \frac{(b^3-1)}{(b^{(3/n)}-1)}$

I am tasked with showing that for any $\epsilon$ there exists $n$ so that $U(f,P_n)-L(f,P_n)<\epsilon$.

I have tried a few approaches. First, I tried

$(b^{(2/n)}-1) (b^{(1/n)}-1) \frac{(b^3-1)}{(b^{(3/n)}-1)}<(b^{(2/n)}) (b^{(1/n)}) (b^3-1)<\epsilon$

to see if that could yield a value of $n$ (dependent on $\epsilon$ and perhaps $b$) but that left me with needing $n$ smaller than some ratio of the logarithms of $b$ and $\epsilon$ which is not necessarily a number above $1$ for all $\epsilon$ (which, again, is required, since $n$ is a positive integer).

Next I tried to directly insert $n=n(\epsilon)$ and see if I get some result smaller than $\epsilon$ but I found nothing that worked well.

Finally I tried the limit of $U(f,P_n)-L(f,P_n)$ as $n$ approaches infinity which equals

$(b^{(0)}-1) (b^{(0)}-1) \frac{(b^3-1)}{(b^{(0)}-1)}=(0)(0) \frac{(b^3-1)}{0}$

Even considering the $0$ in the denominator, this does approach zero, since

$0<U(f,P_n)-L(f,P_n)=(b^{(2/n)}-1) (b^{(1/n)}-1) \frac{(b^3-1)}{(b^{(3/n)}-1)}<(b^{(2/n)}-1) (b^{(1/n)}-1) (b^3-1)$

and the second expression approaches $0$ as $n$ tends towards infinity, so by the Squeeze Theorem the limit of $U(f,P_n)-L(f,P_n)$ must be $0$. (edit: I am very tired. Do I even need this further proof that the limit approaches $0$?)

Therefore, it seems natural to say that since $U(f,P_n)-L(f,P_n) < \epsilon$ for any $\epsilon$ by choosing a sufficiently large $n$ - however, since the limit technically does not exist due to the zero in the denominator, can I really say this?

Or is there perhaps another way entirely to approach this problem?

Answer 1

There shouldn't be any zeros in the denominator. In fact, there shouldn't be any denominators. Let $P = \{x_0, x_1, \ldots, x_n \}$ be a partition of $[1,b]$. Then the upper sum is given by $$U(f,P) = \sum^n_{k=1} M_k (x_k - x_{k-1})$$ where $$M_k = \sup_{x_{k-1} \le x \le x_k} f(x).$$ And the lower sum is the same but with $M_k$ replaced by $$m_k = \inf_{x_{k-1} \le x \le x_k} f(x).$$ For your given partition, we see $$M_k = \sup_{b^{(k-1)/n} \le x \le b^{k/n}} x^2 = b^{2k/n}$$ and $$m_k = \sup_{b^{(k-1)/n} \le x \le b^{k/n}} x^2 = b^{2(k-1)/n}$$ since $x^2$ is an increasing function. Then \begin{align*} U(f,P_n) - L(f,P_n) &= \sum^n_{k=1}(b^{k/n} - b^{(k-1)/n}) b^{2k/n} - \sum^n_{k=1}(b^{k/n} - b^{(k-1)/n}) b^{2(k-1)/n} \\ &= \sum^n_{k=1} (b^{k/n} - b^{(k-1)/n})(b^{2k/n} - b^{2(k-1)/n}) \\ &= \sum^n_{k=1} b^{(k-1)/n}(b^{1/n} - 1) b^{2(k-1)/n}(b^{2/n}- 1) \\ &= \sum^n_{k=1} b^{3(k-1)/n}(b^{1/n} - 1) (b^{2/n} - 1) \\ &\le b^3\sum^n_{k=1} (b^{1/n} - 1) (b^{2/n} - 1) \\ &= b^3 n (b^{1/n} - 1) (b^{2/n} - 1). \end{align*} But by Taylor expansion, we see $$b^{1/n} = e^{\tfrac{\log b}{n}} = 1 + \frac{\log b}{n} + \mathcal O\left(\tfrac 1 {n^2} \right)$$ and similarly $$b^{2/n} = e^{2\tfrac{\log b}{n}} = 1 + \frac{2\log b}{n} + \mathcal O\left(\tfrac 1 {n^2} \right).$$ Then $$(b^{1/n} - 1) (b^{2/n} - 1) = \frac{2 \log(b)^2 }{n^2} + \mathcal O \left( \tfrac 1 {n^3} \right).$$ Then $$U(f,P_n) - L(f, P_n) = \frac{2b^3\log(b)^2}{n} + \mathcal O\left( \tfrac 1 {n^2} \right) \to 0 \,\,\,\,\, \text{ as } n \to \infty.$$ Thus for any $\epsilon > 0$, there is $N \in \mathbb N$ such that $n \ge N$ yields $$U(f,P_n) - L(f, P_n) < \epsilon.$$