**Example 1:** Find a power series representation for the function f(x) = \frac{x}{2x^2+1} and determine the interval of convergence.

**Solution:** Let y = -2x^2 in 1/(1-y)=\sum_{n=0}^\infty y^n. We obtain 1/(1+2x^2)=\sum_{n=0}^\infty (-2x^2)^n = \sum_{n=0}^\infty (-2)^nx^{2n}, and then x/(2x^2+1)=\sum_{n=0}^\infty (-2)^nx^{2n+1}. Because this is a geometric series, it converges when |-2x^2| < 1, that is, |x| < 1/\sqrt{2}. Therefore the interval of convergence is (-1 / \sqrt{2}, 1/\sqrt{2}).

**Example 2:** Find a power series representation and the radius and the interval of convergence for **(a)** f(x) = \ln(x^2+4); **(b)** f(x)=\tan^{-1}(2x).

**Hint: (a)** Write out the power series of f'(x) = 2x/(x^2+4) and integrate both sides; **(b)** Write out the power series of f'(x) = 2/(1+4x^2) and integrate both sides.

**Example 3:** Find the radius of convergence and interval of convergence of the series \sum_{n=1}^\infty \frac{n!x^n}{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}.

**Solution:** Let a_n = \frac{n!x^n}{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}. Note that \frac{a_{n+1}}{a_n} = \frac{(n+1)x}{2n+1} and it goes to x/2 as n\to\infty. By the ratio test, when |x| < 2, the series converges. If |x| = 2, |a_n| = 2\cdot 4\cdot 6\cdot \ldots \cdot (2n) / 1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1) > 1. The limit test says the series diverges. Therefore the radius of convergence is 2 and the interval of convergence is (-2, 2).

**Example 4:** Test the series for convergence or divergence. **(a)** \sum_{n=2}^\infty \frac{1}{n\sqrt{\ln n}}; **(b)** \sum_{n=2}^\infty\frac{(-1)^{n-1}}{\sqrt{n}-1}; **(c)** \sum_{n=1}^\infty\left(\frac{n}{n+1}\right)^{n^2}.

**Hint: (a)** Use the integral test; **(b)** Use the alternating series test; **(c) **Note that \sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}} = \left(\frac{n}{n+1}\right)^{n} = \left(1+1/n\right)^{-n} = e^{-n\ln(1+1/n)}. By l’Hospital’s rule, \lim_{x\to\infty} x\ln(1+1/x) = 1, and so \sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}\to e^{-1} < 1 as n\to\infty. Use the root test.