Let \[f_n(x) = \begin{cases} \frac{1}{n} & x \in (\frac{1}{2^{n+1}}, \frac{1}{2^n}] \\ 0 & \text{ otherwise}.\end{cases}\]
Show that \(\sum_{n=1}^\infty f_n\) does not satisfy the Weierstrass M-test but that it nevertheless converges uniformly on \(\mathbb{R}\).
Solution
1. The Weierstrass M-test does not apply.
the M-test requires \(\sum_n M_n < \infty\) where \(M_n = \sup_x |f_n(x)|\).
2. The supports \(\supp f_n = (2^{-(n+1)}, 2^{-n}]\) are pairwise disjoint.
the intervals \((2^{-(n+1)}, 2^{-n}]\) for \(n = 1, 2, \ldots\) are disjoint (they partition \((0, 1/2]\)), and \(f_n = 0\) elsewhere.
3. Pointwise limit: \(S(x) = \sum_n f_n(x) = f_{n(x)}(x)\) where \(n(x)\) is the unique index with \(x \in (2^{-(n+1)}, 2^{-n}]\), and \(S(x) = 0\) otherwise.
by 2, at most one term is nonzero at each \(x\), so the series has at most one nonzero summand and converges.
4. Uniform convergence on \(\RR\).
for the tail, \(S(x) - S_N(x) = \sum_{n>N} f_n(x)\), which by 2 is either \(0\) (if the support containing \(x\) has index \(\le N\)) or \(f_n(x) = 1/n \le 1/(N+1)\) (if its index is \(n > N\)). Hence \[ \sup_{x\in\RR}\big|S(x) - S_N(x)\big| \le \frac{1}{N+1} \to 0, \] so the series converges uniformly.
5. Q.E.D.