T.4 · Sequences, series and uniform convergence

A convergent real sequence is Cauchy, and every real Cauchy sequence converges because R is complete. In Q a Cauchy sequence can approach an irrational number and fail to converge within Q. A bounded monotone real sequence converges. A bounded sequence need not converge, but Bolzano–Weierstrass guarantees a convergent subsequence; (-1)^n has two different subsequential limits.

For a numerical series, absolute convergence implies convergence; conditional convergence does not allow arbitrary rearrangement without affecting the sum. The alternating harmonic series converges but its absolute-value series diverges. In a ratio test, a limit below 1 proves absolute convergence and one above 1 proves divergence; a limit equal to 1 is inconclusive, as both sum 1/n and sum 1/n² demonstrate.

Pointwise convergence chooses an index N separately for each x and tolerance. Uniform convergence chooses one N that works for every x in the domain. For real-valued functions, check the supremum of |f_n−f| over the whole domain. A continuous pointwise limit does not by itself prove uniform convergence. Domain endpoints and shrinking peaks often distinguish the two notions.

A uniform limit of continuous functions is continuous. On a closed bounded interval, uniform convergence of Riemann-integrable functions permits exchanging limit and integral. Exchanging derivatives needs extra hypotheses; uniform convergence of the functions alone is insufficient. The Weierstrass M-test establishes uniform absolute convergence of a series of functions if each term is bounded by M_n on the whole domain and sum M_n converges.

On [0,1], f_n(x)=x^n tends to 0 for x<1 and to 1 at x=1. The limit is discontinuous, so convergence cannot be uniform; directly, the supremum error is 1, approached below 1. On [0,a] with 0≤a<1, the limit is zero and the supremum is a^n, which tends to zero, so convergence is uniform. Changing the domain changes the conclusion.

Checking several fixed x values proves neither a supremum bound nor uniform convergence. The point producing the largest error may move with n.

Determine pointwise and uniform convergence of $f_n(x)=x^n$ on $[0,1]$ and on $[0,a]$ for fixed $0. Compare $\sum1/n^2$ with the alternating harmonic series for absolute convergence.

Worked solution

On $[0,1]$, the limit is zero for $x<1$ and one at 1. This limit is discontinuous, so continuous $f_n$ cannot converge uniformly. More directly the supremum error on $[0,1)$ is one. On $[0,a]$ the supremum is $a^n\to0$, so convergence is uniform to zero. The p-series with p=2 converges absolutely. The alternating harmonic series converges by decreasing terms tending to zero, but the absolute series diverges; its convergence is conditional.

Let $g_n(x)=\sin(nx)/n$ on $\mathbb R$. Is convergence uniform? May its derivatives converge to the derivative of its limit? Use an explicit point to decide.

Check after attempting

$\sup_{x\in\mathbb R}|g_n(x)|=1/n\to0$, so the functions converge uniformly to zero. But $g_n'(x)=\cos(nx)$, and at $x=0$ these derivatives are all 1. The derivative of the zero limit is 0. Thus uniform convergence of the functions alone does not justify differentiating a limit; additional derivative-convergence hypotheses are needed.