Integration sans fundamental theorem of calculus?

Question

Hope you all had a Merry Christmas yesterday. The challenge du jour:

Let n be an arbitrary fixed positive integer, and consider the right-hand Riemann sums of the function x^n obtained from a partition of [0, 1] into m equal intervals. These are:

[k=1, m]∑(1/m * (k/m)^n)

Using the fundamental theorem of calculus, it is of course trivial to prove that:

[m→∞]lim [k=1, m]∑(1/m * (k/m)^n) = [0, 1]∫x^n dx = x^(n+1)/(n+1) |[0, 1] = 1/(n+1).

Your mission, should you decide to accept it, is to prove the limit [m→∞]lim [k=1, m]∑(1/m * (k/m)^n) = 1/(n+1) _directly_, without using the fundamental theorem. Can you do it?

Okay, then. How do you do it? Smartass.

ksoileau: "It follows trivially from the fact that

[m→∞]lim [k=1, m]∑(1/m * (k/m)^(1/n)) =n/(n+1)."

And how do you prove that fact without using the FTC?

Steiner: Wonderful. This is indeed the method which I was thinking of. However, I would like to see you expound further on the assertion that "1^n +2^n....+ k^n is a polynomial P of degree n +1, in k, with leading term 1/(n +1)." How is this proven, for a general positive integer n? You don't have to worry about fractional n, since I specified in the original post that n is a positive integer.

To clarify, I would like a direct proof that the first term in P is 1/(n+1), not just an appeal to Faulhaber's formula (unless you wish to prove Faulhaber's formula in the post? Although it's probably easier to just focus on the first term).

Answer 1

Hope you had a Merry Christmas too and have a happy 2008. In case someone doesn't know,

sans -- without
du jour - of the day (By the way, challenge -noun - in French is défi. Défi du jour - challenge of the day


Is n a positive integer? If so, it is easy. The conclusion follows from the fact that 1^n +2^n....+ k^n is a polynomial P of degree n +1, in k, with leading term 1/(n +1). So, your limit is the same as lim (k --> oo) P(k)/k^(n+1) = 1/(n +1).

But if n is not an integer, then , well, at least for now I don't have the answer. I thought about using the Lebesgue integral with the Lebesgue measure. Your sequence can be seen as a sequence of integrals of simple functions and these integrals will converge to the Lebesgue integral of x^n over [0,1]. If you can compute this Lebesgue integral directly, without using the fact that it coincides with the function's Riemann integral, then you don't need the fundamental theorem of calculus.

Edit:

Ok, so let's suppose n is a positive integer. For every k =1,2,3... we have, according to the binomial expansion,

(1 + k)^(n+1) = k^(n +1) +(n +1) k^n + [(n+1)n]/2 k^(n-1) ...+ 1

(1 + k-1)^(n +1) = (k-1)^(n +1) +(n +1) (k-1)^n + [(n+1)n]/2 (k-1)^(n-1) ...+ 1
.
.
(1 +1)^(n+1) = (1)^(n +1) +(n +1) (1)^n + [(n+1)n]/2 (1)^(n-1) ...+ 1

We observe that the first term in the right hand side of each equality, except the term of the last one, is just the term in the left hand side of the equality immediately below. So, if we add these k inequalities, we get

(1 + k)^(n+1) = 1 + (n+1)[k^n + (k-1)^n...+1^n] + ((n+1)n/2) [k^(n-1) + (k-1)^(n-2) ...+1^(n-1)]....

So,
k^n + (k-1)^n...+1^n = 1/(n+1) {(1+k)^(n+1) - ((n+1)n/2) [k^(n-1) + (k-1)^(n-2) ...+1^(n-1)]....... -1}.

This provides a recurrence formula that allows us to determine the sum oh the nth powers of 1,2,...k knowing such sum for the powers n-1, n-2,......1. It requires a lot of algebra, but, for our purpose, the important thing is this shows that, if the sums for n-1, n-2...1 are polynomials in k of degree n, n-1, ...2, then the sum for the power n is a polynomial in k of degree n +1. This also show very clearly that, supposing this is the case, then the leading coefficient of the polynomial which gives the sum of powers n is 1/(n+1).

We know that 1^1 + 2^1 +....k^1 = k(k+1)/2, a polynomial of degree 1+1 = 2. this is just the sum of the k first terms of an arithmetic progression. So, by induction on n, we have the desired proof

In fact

1 + 2 ...+ k = k(k+1)/2 - leading coefficient is 1/2
1^2 + 2^2 .+ k^2 = k(k+1)(2k+1)/6 - leading coefficient is 1/3
1^3 + 2^3 ....+k^3 = [k(k+1)/2]^2 - leading coefficient is 1/4

From 4 on, I don't know by heart the corresponding formulas.....

Happy 2008!

Answer 2

It follows trivially from the fact that

[m→∞]lim [k=1, m]∑(1/m * (k/m)^(1/n)) =n/(n+1).

Answer 3

So your question is can i do it? Well then my answer is simply yes! =p

Wheher i want to write it down... well.