Preface

This is a problem from Courant and Robbins'What is Mathematics?, Chapter 8: "The Calculus", page 409-410.

Also, I apologize if the format of this post leaves much to be desired. I've never used LaTex before, so I patched this together as best as I was able to. Helpful suggestions to improve in that direction would likewise be welcome.

Request

I wouldloveany helpful hints/nudges in the right direction, but I would prefer not to have the solution spelled out for me just yet.

The Problem

Prove that for any rational $\displaystyle k \neq -1$ the same limit formula, $\displaystyle N \rightarrow k+1$, and therefore the result:

$\displaystyle \int_{a}^{b} x^{k} dx = \frac {b^{k+1} - a^{k+1}} {{k+1}} $, k any positive integer

remains valid. First give the proof, according to our model, for negative integers k.Then, if $\displaystyle k = \frac {u}{v}$, write $\displaystyle q^{\frac{1}{v}} = s$ and

$\displaystyle N = \frac{s^{k+1} -1}{s^{v} -1} = \frac{s^{u+v} -1}{s^{v} -1} = \frac{s^{u+v} -1}{s -1} / \frac{s^{v} -1}{s -1}$

Relevant Information

This was the proof that preceded it.

To obtain the integral formula

$\displaystyle \int_{a}^{b} x^{k} dx = \frac {b^{k+1} - a^{k+1}} {{k+1}}$

we form $\displaystyle S_{n}$ by choosing the points of subdivision $\displaystyle x_{0} = a, x_{1}, x_{2}, ... , x_{n} = b$ in geometrical progression. We set the $\displaystyle ^{n}\sqrt{\frac{b}{a}} = q$ , so that $\displaystyle \frac{b}{a} = q^{n}$ , and define $\displaystyle x_{0} = a, x_{1} = aq, x_{2} = aq², ... , x_{n} = aq^{n} = b$ .

By this device, as we shall see, the passage to the limit becomes very easy. For the "rectangle sum" $\displaystyle S_{n}$ we find, since $\displaystyle f(x_{i}) = x^{k}_{i} = a^{k}q^{ik}$ , and $\displaystyle \Delta x = x_{i+1} - x_{i} = aq^{i+1} - aq^{i}$ ,

$\displaystyle S_{n} = a^{k}(aq - a) + a^{k}q^{k}(aq^{2} - aq) +a^{k}q^{2k}(aq^{3} - aq^{2}) + ... + a^{k}q^{k(n-1)}(aq^{n} - aq^{n-1})$

Since each term contains the factor $\displaystyle a^{k}(aq - a)$ we may write,

$\displaystyle S_{n} = a^{k+1}(q - 1)[1 + q^{k+1} + q^{2(k+1)} + ... + q^{(n-1)(k+1)}]$

Substituting $\displaystyle t$ for $\displaystyle q^{k+1}$ we see that the expression in braces is the geometrical series$\displaystyle 1 + t + t^{2} + ... + t^{(n-1)}$ , whose sum is $\displaystyle \frac{t^{n} - 1}{t -1}$ . But $\displaystyle t^{n} = q^{n(k+1)} = (\frac {b}{a})^{k+1} = \frac{b^{k+1}}{a^{k+1}}$ . Hence,

$\displaystyle S_{n} = (q - 1) [\frac {b^{k+1} - a^{k+1}}{q^{k+1} - 1}] = \frac {b^{k+1} - a^{k+1}}{N}$

where $\displaystyle N = \frac{q^{k+1} -1}{q -1}$

Thus far, n has been a fixed number. Now we shall let n increase, and determine the limit of N. As n increases, the $\displaystyle ^{n}\sqrt{\frac{b}{a}} = q$ will tend to 1, and therefore both numerator and denominator of N will tend to zero, which makes caution necessary. Suppose first that k is a positive integer; then the division of q -1 can be carried out, and we obtain $\displaystyle N = q^{k} + q^{k-1} + ... + q + 1$ . If now n increases, q tends to 1 and hence $\displaystyle q^{2}, q^{3}, ... , q^{k}$ will also tend to 1, so that N approaches k+1. But this shows that $\displaystyle S_{n}$ tends to $\displaystyle \frac {b^{k+1} - a^{k+1}} {{k+1}}$ , which was to be proved.

Attempts / Ideas

I'm not certain where I should even start. I thought that, perhaps, simply following in their steps and approaching the problem as a geometrical series would yield results; the notable difference being that instead of $\displaystyle a^{k}q^{ik}$ we would have $\displaystyle a^{-k}q^{-ik}$, giving us a geometrical series that looks, eventually, like $\displaystyle 1 + \frac{1}{t} + \frac{1}{t^{2}} + ... + \frac{1}{t^{n+1}}$. $\displaystyle \Delta x$ remains the same.

This did not appear to get me anywhere, though I freely admit that, perhaps, it got me somewhere and I simply did not realize it. At this point, for the sake of my sanity and my slumber, any help that points me in the right direction would be very much appreciated.