Discussion
For which one of the following pairs of values of \(a\) and \(b\), does the series \(\sum_{k=1}^{\infty} \frac{(3x)^{k}}{2^{\sqrt{k}}(1-x)^{k}}\) converge for all \(x \in (a,b)\)?
Answer: C
The pair is \(a = \frac{-1}{2}\) and \(b = \frac{1}{4}\).
Write \(t = \frac{3x}{1-x}\), so the term is \(\frac{t^{k}}{2^{\sqrt{k}}}\). Since \((2^{\sqrt{k}})^{1/k} = 2^{1/\sqrt{k}} \to 1\), the root test gives convergence for \(|t| \lt 1\) and divergence for \(|t| \gt 1\). At \(|t| = 1\) the terms are \(\pm 2^{-\sqrt{k}}\), which are summable because \(2^{\sqrt{k}}\) eventually exceeds \(k^{2}\). So the series converges exactly when \(|3x| \leq |1 - x|\):
\[ \begin{aligned} 9x^{2} &\leq (1-x)^{2} \\ 8x^{2} + 2x - 1 &\leq 0 \\ (4x - 1)(2x + 1) &\leq 0 \\ -\tfrac{1}{2} \leq x &\leq \tfrac{1}{4} \end{aligned} \]
The interval \((a,b)\) must sit inside \([-\frac{1}{2}, \frac{1}{4}]\), and \((-\frac{1}{2}, \frac{1}{4})\) does. The others reach outside it: \(b = \frac{1}{2}\) and \(b = \frac{1}{3}\) include \(x = 0.3\), where \(t = \frac{9}{7} \gt 1\); \(a = -1\) includes \(x = -0.9\), where \(|t| = \frac{27}{19} \gt 1\).
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