Start from a series you know, and re-check the interval BC only

These four cover almost everything the course asks for:

1. $\displaystyle\frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} = 1 + x + x^{2} + \cdots$, for $|x| < 1$.
2. $\displaystyle e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \cdots$, for all $x$.
3. $\displaystyle \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} = x - \frac{x^{3}}{3!} + \cdots$, for all $x$.
4. $\displaystyle \cos x = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n}}{(2n)!} = 1 - \frac{x^{2}}{2!} + \cdots$, for all $x$.

Two checks that catch a misremembered one in seconds, and they are worth using rather than trusting recall:

1. Check the parity. $\sin$ is odd, so only odd powers; $\cos$ is even, so only even powers. Any recalled series with the wrong parity is wrong before any coefficient is examined.
2. Check a value. $\cos 0 = 1$, so the series must have a constant term of $1$. $\sin 0 = 0$, so it must have none. $e^{0} = 1$. Each of these kills a whole family of misremembering.

The geometric one is different in kind: it is not memorised so much as derived, straight from 10.2 with $a = 1$ and $r = x$. That also explains its interval, since $|r| < 1$ is exactly the geometric condition. If the other three ever desert you, this one can be rebuilt from first principles.

Anywhere $x$ appears you may substitute another expression, provided you substitute it into the condition as well. That second half is the one that gets dropped.

1. $\frac{1}{1+x}$: substitute $-x$, giving $\sum (-1)^{n}x^{n}$. The condition $|-x| < 1$ is still $|x| < 1$, so the interval is unchanged.
2. $\frac{1}{1-x^{2}}$: substitute $x^{2}$, giving $\sum x^{2n}$. The condition is $|x^{2}| < 1$, which is still $|x| < 1$ but for a different reason.
3. $e^{-x^{2}}$: substitute $-x^{2}$ into the exponential series, giving $\sum \frac{(-1)^{n}x^{2n}}{n!}$, convergent for all $x$. Note the factorial stays $n!$, since it came from the original series' index rather than from the new exponent.
4. $\frac{1}{1-2x}$: substitute $2x$, giving $\sum 2^{n}x^{n}$. Now $|2x| < 1$, so $|x| < \frac{1}{2}$ and the radius has halved.

That last case is the one to remember, because it is the case where the interval genuinely changes and where forgetting to substitute into the condition produces an interval twice as wide as the truth.

Inside its interval of convergence a power series may be differentiated or integrated one term at a time, and the result is the series of the derivative or antiderivative. The radius is unchanged by both operations. The endpoints are not.

The standard derivation, and the standard trap, in one example. Starting from $\frac{1}{1+x} = \sum (-1)^{n}x^{n}$ on $(-1, 1)$ and integrating:

$$\ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{n+1}}{n+1} = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots.$$

The radius is still $1$. But at $x = 1$ the new series is the alternating harmonic series, which converges, while the original series at $x = 1$ was $\sum (-1)^{n}$, which does not. Integration gained an endpoint. The interval went from $(-1, 1)$ to $(-1, 1]$.

Two further notes:

1. Integration introduces a constant. Determine it by substituting a convenient value, usually the centre. Here $\ln(1+0) = 0$ fixes $C = 0$.
2. Differentiation can lose an endpoint, by the same argument run backwards. Differentiating $\sum \frac{x^{n}}{n^{2}}$, convergent on $[-1, 1]$, gives $\sum \frac{x^{n-1}}{n}$, which no longer converges at $x = 1$.

Every operation in this topic is safe, and every one of them changes something. The discipline is one line, applied after each step:

1. What did this do to the radius? Substitution can change it; differentiation and integration cannot.
2. What did this do to the endpoints? Integration can gain one; differentiation can lose one; substitution replaces them entirely.

Multiplying by a power of $x$ is the gentlest operation: $x^{3}\sum a_n x^{n} = \sum a_n x^{n+3}$ shifts every exponent and changes neither the radius nor the endpoints, because a nonzero constant multiple of a term does not affect convergence.

And one caution about substitution that outranks the rest. Substituting an expression that leaves the original interval is not a valid operation at all. The series for $\frac{1}{1-x}$ holds for $|x| < 1$; substituting $x = 2$ produces $\sum 2^{n}$, which diverges, and the fact that $\frac{1}{1-2} = -1$ is a finite number does not rescue it. That is 10.2's negative-sum warning returning as the last idea in the course.