Determine the values of \(x\in\mathbb{R}\) for which \(\displaystyle\sum_{n=1}^\infty \frac{x^n}{1+n|x|^n}\) converges, justifying your answer carefully.
Solution
1. For \(|x| < 1\): the series converges absolutely.
\(\left|\dfrac{x^n}{1 + n|x|^n}\right| \le |x|^n\) (denominator \(\ge 1\)), and \(\sum |x|^n\) is a convergent geometric series.
2. At \(x = 1\): the series diverges to \(+\infty\).
every term is positive and \(\dfrac{1}{1+n} \ge \dfrac{1}{2n}\), so the series dominates a divergent harmonic tail.
3. At \(x = -1\): the series converges.
\(\sum \dfrac{(-1)^n}{1+n}\) is an alternating series whose terms decrease to \(0\); Leibniz’s test applies.
4. For \(x > 1\): the series diverges to \(+\infty\).
every term is positive and \(\dfrac{x^n}{1 + nx^n} = \dfrac{1}{n + x^{-n}} \ge \dfrac{1}{n+1}\), so the series dominates \(\sum \dfrac{1}{n+1} = \infty\).
5. For \(x < -1\): the series converges (conditionally).
5.1. Write the terms as \((-1)^n a_n\) with \(a_n = \dfrac{|x|^n}{1 + n|x|^n} = \dfrac{1}{n + |x|^{-n}}\).
\(x^n = (-1)^n|x|^n\); divide numerator and denominator by \(|x|^n\).
5.2. \(a_n \downarrow 0\).
\(n + |x|^{-n} \uparrow \infty\); the increment \(\big((n+1) + |x|^{-(n+1)}\big) - \big(n + |x|^{-n}\big) = 1 - |x|^{-n}(1 - 1/|x|) > 0\) since \(|x|^{-n}(1 - 1/|x|) < 1\) for \(|x| > 1\).
5.3. \(\sum (-1)^n a_n\) converges.
5.2.
5.4. The convergence is not absolute.
\(a_n \ge \dfrac{1}{n+1}\) (by <2>1), so \(\sum a_n\) diverges.
6. Q.E.D.: the series converges for \(x \in (-\infty, 1)\) — absolutely for \(|x| < 1\), conditionally for \(x \le -1\) — and diverges for \(x \ge 1\).
<1>1–<1>5 cover all real \(x\).