Find all real numbers \(x\) for which the series \(\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n^2+1}}\) converges. [Be sure to justify your analysis.]
Solution
Consider \[ \sum_{n=1}^\infty \frac{x^n}{\sqrt{n^2+1}}. \] If \(|x|<1\), then \[ 0\le \frac{|x|^n}{\sqrt{n^2+1}}\le |x|^n, \] so the series converges absolutely by comparison with a geometric series.
If \(|x|>1\), then \[ \frac{|x|^n}{\sqrt{n^2+1}} \sim \frac{|x|^n}{n}, \] which does not tend to \(0\); hence the series diverges.
At \(x=1\), \[ \frac1{\sqrt{n^2+1}}\ge \frac1{\sqrt2\,n}, \] so the series diverges by comparison with the harmonic series.
At \(x=-1\), the series is \[ \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+1}}. \] The positive terms \(1/\sqrt{n^2+1}\) decrease to \(0\), so the alternating-series test gives convergence.
Therefore the set of real \(x\) for which the series converges is \[ [-1,1). \]