Upper and Lower Sums Close In

Use $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ to prove $\int_0^1 x^2\,dx = \frac13$ from upper and lower sums.

L_n = \frac{(n-1)n(2n-1)}{6n^3} \to \frac13, \quad U_n = \frac{n(n+1)(2n+1)}{6n^3} \to \frac13

n = 48

: the gap

U - L = 1/48 \approx 0.0208

squeezes

\frac13

.

Show proof

On the uniform partition $P_n$ , $x^2$ is increasing, so the lower sum uses left endpoints and the upper sum right endpoints: $L_n = \frac{1}{n^3}\sum_{k=0}^{n-1}k^2$ , $U_n = \frac{1}{n^3}\sum_{k=1}^{n}k^2$ .
Substituting the sum-of-squares formula gives $L_n = \frac{(n-1)n(2n-1)}{6n^3}$ and $U_n = \frac{n(n+1)(2n+1)}{6n^3}$ .
Both tend to $\frac{2n^3}{6n^3} = \frac13$ as $n \to \infty$ , so the integral is $\frac13$ by squeezing.

Prove the Dirichlet function is not Riemann integrable by showing $U - L = 1$ for every partition.

U(D,P) = 1, \quad L(D,P) = 0 \quad \text{for every } P

Contrast with the sums for

x^2

— for the Dirichlet function the gap never narrows.

Show proof

Every interval contains both rationals and irrationals, so on every subinterval $\sup D = 1$ and $\inf D = 0$ .
Hence for every partition $P$ , $U(D,P) = \sum 1\cdot\Delta x_k = 1$ and $L(D,P) = \sum 0\cdot\Delta x_k = 0$ .
Since $U - L = 1 \not\to 0$ , the necessary and sufficient condition for Riemann integrability fails.

Prove that refining a partition never increases $U - L$ .

P' \supset P \;\Rightarrow\; U(f,P') \le U(f,P),\; L(f,P') \ge L(f,P)

n = 10

: the gap shrinks monotonically as the partition refines.

Show proof

Splitting one subinterval in two: each piece's supremum is at most the original's, so the upper sum decreases (or stays).
Likewise each piece's infimum is at least the original's, so the lower sum increases (or stays).
So $U - L$ decreases monotonically under refinement — the sums squeeze toward each other.

Part of these learning paths