Show that the infinite series \(\sum_{n=0}^\infty x^n2^{-nx}\) converges uniformly on \([0,B]\) for any \(B > 0\). Does this series converge uniformly on \([0,\infty)\)?
Solution
1. The terms are \(x^n2^{-nx} = (x2^{-x})^n\).
\(2^{-nx} = (2^{-x})^n\).
2. \(\sup_{x \ge 0} x2^{-x} = \frac{1}{e\ln 2} < 1\).
2.1. The function \(h(x) = x2^{-x} = xe^{-x\ln 2}\) has its global maximum on \([0,\infty)\) at \(x = 1/\ln 2\).
\(h'(x) = e^{-x\ln 2}(1 - x\ln 2)\) vanishes only at \(x = 1/\ln 2\), where \(h\) changes from increasing to decreasing; \(h(0) = 0\) and \(h \to 0\) as \(x \to \infty\).
2.2. \(h(1/\ln 2) = \frac{1}{e\ln 2} \approx 0.531 < 1\).
\(h(1/\ln 2) = \frac{1}{\ln 2}e^{-1} = \frac{1}{e\ln 2}\), and \(e\ln 2 > 1\).
3. The series converges uniformly on \([0, B]\) for every \(B > 0\).
on \([0,B]\), \(x2^{-x} \le \max(h(B), h(1/\ln 2)) < 1\) by 2, so the Weierstrass M-test applies: \(\sum (x2^{-x})^n \le \sum r^n\) with \(r < 1\).
4. The series converges uniformly on \([0, \infty)\) as well.
2 gives a uniform ratio bound \(x2^{-x} \le \frac{1}{e\ln 2} < 1\) on all of \([0,\infty)\), so the M-test applies globally: \(\left|\sum_{n>N}(x2^{-x})^n\right| \le \frac{r^{N+1}}{1-r}\) with \(r = \frac{1}{e\ln 2}\), which \(\to 0\) as \(N \to \infty\) independent of \(x\).