(June 2009 #4b) Prove that the series \[x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\frac{x^2}{(1+x^2)^3}+\cdots\] converges uniformly on \([a,\infty)\) for every \(a>0\); but not uniformly on \([0,b]\) for any \(b>0\).
Solution
Goal: Prove \(\sum_{k=0}^\infty \frac{x^2}{(1+x^2)^k}\) converges uniformly on \([a, \infty)\) for each \(a > 0\) but not uniformly on \([0, b]\) for any \(b > 0\).
1. The series is geometric with ratio \(1/(1+x^2)\): \(\sum_{k=0}^\infty \frac{x^2}{(1+x^2)^k} = x^2 \sum_{k=0}^\infty \left(\frac{1}{1+x^2}\right)^k\).Proof: factor \(x^2\) out of every term.
2. Pointwise sum: \(S(x) = \lim_N \sum_{k=0}^N \frac{x^2}{(1+x^2)^k}\) equals \(1 + x^2\) for \(x \neq 0\) and \(0\) at \(x = 0\).Proof: for \(x \neq 0\), the ratio \(1/(1+x^2) \in (0,1)\) and the geometric formula gives \(x^2 \cdot \frac{1}{1 - 1/(1+x^2)} = x^2 \cdot \frac{1+x^2}{x^2} = 1 + x^2\).At \(x = 0\) every term is \(0\).
3. The \(N\)-th tail is \(R_N(x) := \sum_{k=N+1}^\infty \frac{x^2}{(1+x^2)^k} = \frac{1}{(1+x^2)^N}\) for \(x \neq 0\), and \(R_N(0) = 0\).Proof: geometric tail: \(\frac{x^2}{(1+x^2)^{N+1}} \cdot \frac{1}{1 - 1/(1+x^2)} = \frac{x^2}{(1+x^2)^{N+1}} \cdot \frac{1+x^2}{x^2} = \frac{1}{(1+x^2)^N}\).
4. Uniform convergence on \([a, \infty)\) for \(a > 0\): \(\sup_{x \ge a} |R_N(x)| \le \frac{1}{(1+a^2)^N} \to 0\).Proof: for \(x \ge a\), \(1 + x^2 \ge 1 + a^2\), so \(R_N(x) = (1+x^2)^{-N} \le (1+a^2)^{-N}\) by 3; and \((1+a^2)^{-N} \to 0\) since \(1 + a^2 > 1\).
5. No uniform convergence on \([0, b]\) for \(b > 0\): \(\sup_{x \in [0,b]} |R_N(x)| = 1\) for every \(N\).Proof: by 3, \(R_N(0) = 0\) but \(R_N(x) = (1+x^2)^{-N} \to 1\) as \(x \to 0^+\); in fact \(\sup_{x \in (0,b]} (1+x^2)^{-N} = 1\) (approached as \(x \to 0\)). So the suprema of the tails do not tend to \(0\), which is exactly the negation of uniform convergence.(Equivalently, the pointwise limit \(S\) of 2 is discontinuous at \(0\) while partial sums are continuous.)
6. Q.E.D. Proof: 4 and 5 are the two claims.