Representing Functions as Power Series BC only

Four Maclaurin series carry most of the course: $\frac{1}{1-x} = \sum x^{n}$ for $|x| < 1$, $e^{x} = \sum \frac{x^{n}}{n!}$, $\sin x = \sum \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ and $\cos x = \sum \frac{(-1)^{n}x^{2n}}{(2n)!}$, the last three for all $x$. A misremembered one is caught by parity, since $\sin$ has only odd powers and $\cos$ only even, or by checking a known value, since $\cos 0 = 1$ forces a constant term and $\sin 0 = 0$ forbids one. The geometric series is derived rather than memorised, straight from 10.2 with $r = x$, which is also where its interval comes from.

Substitution must be made into the condition as well as the series: $x^{2}$ leaves $|x| < 1$ intact, while $2x$ gives $|2x| < 1$ and halves the radius. Term-by-term differentiation and integration are legal inside the interval and leave the radius unchanged, but not the endpoints: integrating $\frac{1}{1+x} = \sum(-1)^{n}x^{n}$ gives $\ln(1+x) = \sum \frac{(-1)^{n}x^{n+1}}{n+1}$, whose value at $x = 1$ is the convergent alternating harmonic series where the original had the divergent $\sum(-1)^{n}$, so the interval gained an endpoint and became $(-1, 1]$; differentiation can lose one by the same argument reversed. Integration also introduces a constant, fixed by substituting the centre. Multiplying by a power of $x$ shifts every exponent and changes nothing about convergence, while substituting a value outside the original interval is not an operation at all, since $\sum 2^{n}$ diverges however finite $\frac{1}{1-2}$ may look.