Question

Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of convergence. $$ g(x)=\frac{1}{x^{2}+1} $$

Short answer

The function \(g(x)=\frac{1}{x^{2}+1}\) can be represented by the power series \(\sum_{n=0}^{\infty} x^{2n}\), and the interval of convergence is \(-1 \le x \le 1\).

Step-by-step solution

Replace \(x\) in the known series

Start by inserting \(-x^{2}\) for \(x\) in the known series for \(\frac{1}{1+x}\): \[g(x) = \sum_{n=0}^{\infty}(-1)^{n} (-x^{2})^{n} = \sum_{n=0}^{\infty}(-1)^n (-1)^n x^{2n}\]

Simplify the series

Simplify the series using the fact that \((-1)^{n} \cdot (-1)^{n} = 1^{2n}=1\): \[g(x) = \sum_{n=0}^{\infty} x^{2n}\]

Determine the interval of convergence

Since the original series converged for \(-1< x <1\), now, replacing \(x\) with \(-x^{2}\), we have \(-1 < -x^{2} < 1\). That leads to \(1 > x^{2} > -1\). Square rooting all components yield \(-1 < x < 1\). Thus, the interval of convergence remains the same. Note that the series does converge at \(x=\pm 1\), so the convergence interval is \(-1 \le x \le 1\).