# Infinite series

http://www.physics.miami.edu/~nearing/mathmethods/series.pdf

## Series worth knowing

Geometric Series
$\sum\limits_0^\infty x^n = 1 + x + x^2 + x^3 + \ldots = \frac{1}{1-x}$
Consider: $(1 + x + x^2 + \ldots + x^N)(1-x) = 1 - x^{N+1}$
so $\sum\limits_0^N x^N = \frac{1-x^{N+1}}{1-x}$

Power Series

• $e^x = 1 + x + \frac{x^2}{2!} + \ldots = \sum\limits_0^\infty \frac{x^k}{k!}$
• $\sin x = x - \frac{x^3}{3!} + \ldots = \sum\limits_0^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$
• $\cos x = 1 - \frac{x^2}{2!} + \ldots = \sum\limits_0^\infty (-1)^k \frac{x^{2k}}{(2k)!}$
• $\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots = \sum\limits_1^\infty (-1)^{k+1} \frac{x^k}{k}$ for $|x| < 1$
• $(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)x^2}{2!} + \ldots = \sum\limits_0^\infty \frac{\alpha(\alpha-1) \ldots (\alpha - k + 1)}{k!} x^k$ for $|x| < 1$ (Binomial series)

Taylor Series
$\displaystyle \begin{array}{rl} f(x) &= f(0) + xf'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \ldots \\ &= f(t_0) + (t - t_0)f'(t_0) + \frac{(t - t_0)^2}{2!} f''(t_0) + \ldots \end{array}$

## Convergence Tests

Comparison test
Let $u_k, v_k$ be sequences of positive reals and $\exists k_0 < \infty$ s.t. $\forall_{k < k_0} u_k < v_k$ and $v_k$ converges, then $u_k$ converges

Ratio test
If for large $k$, $\frac{u_{k+1}}{u_k} \leq x \le 1$, then $\sum u_k$ converges. Derived from comparison test with geometric series.

Integral test
If $f(x)$ is decreasing positive and $\int_0^\infty f(x) dx$ converges, $\sum\limits_0^\infty f(x)$ converges

## Stirling’s approximation

$\displaystyle n! = \Gamma (n + 1) = \int_0^\infty t^n e^{-t} dt = \int_0^\infty e^{-t + n\ln t}$

Using newton’s method to approximate this to the 2nd order:

$\displaystyle \Gamma (n + 1) \sim n^n e^{-n} \int_{-\infty}^\infty e^{-(t-n)^2 / 2n} dt = n^n e^{-n} \sqrt{2\pi n}$ (for moderately large n)

The last step in the above makes use of the Gaussian integral:

$\displaystyle \int_{-\infty}^\infty e^{- \alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$

using parametric differentiation:

$\displaystyle \int_{-\infty}^\infty x^2 e^{-\alpha x^2} dx = -\frac{d}{d\alpha} \sqrt{\frac{\pi}{\alpha}} = \frac{1}{2} \left( \frac{\sqrt{\pi}}{\alpha^{3/2}} \right)$