Power Series

A series

\sum_{n = 0}^{\infty} b_{n} (x - a)^{n}

Where x is a variable, $b_{n}$ a real-valued coefficient, and the series is centered about $x = a$

If $x = a,$ all terms but the first one are zero, so the series always converges $x = a$

A power series may converge for some $x$ 's and diverge for others

As $x = a$ always converges, we can find a symmetric radius $R$ around $x = a$ such that a power series converges if $| x - a | < R$ and diverges for $| x - a | > R$

$R is called the radius of convergence$

All values of $x$ within $R$ units of $x = a$ produce a convergent series

| x - a | < R \to a - R < x < a + R

The Ratio Test is often used to find $R$ since it produces a relationship involving $| x - a |$

Convergence will occur when the limit in the ratio test is $< 1,$ and the limit will involve $x$

The interval of convergence consists of all values of $x$ within the radius of convergence and will always be at least $(a - R, a + R)$ but may also include one or both endpoints of this interval

Ratio Test in inconclusive for $lim_{n \to \infty} = 1$ , so it does not provide information about the endpoints of the interval
needs to be manually checked

Knowing that $\sum a r^{n}$ converges to $\frac{a}{1 - r}$ , we can manipulate the function $\frac{1}{1 - x}$ to build power series to other functions

\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n}, | x | < 1

f (x) = \frac{x^{3}}{1 + 4 x^{2}} = \frac{1}{1 - (- 4 x^{2})} \cdot x^{3} = \sum_{n = 0}^{\infty} (- 4 x^{2})^{n} x^{3}, | - 4 x^{2} | < 1 \to \sum_{n = 0}^{\infty} (- 1)^{n} 4^{n} x^{2 n + 3}, | x | < \frac{1}{2}

We can also integrate, differentiate, etc to manipulate the functions