Write four terms, then find the general one BC only

The Taylor series for $f$ centred at $a$ is the polynomial of 10.11 with no upper limit:

$$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^{n}.$$

Everything from 10.11 carries over unchanged: the derivative order, the factorial and the exponent are still the same number in every term, and the centre still appears twice. Two things are new:

1. You must produce a general term, valid for every $n$, rather than stopping at a chosen degree.
2. Convergence is now a question. An infinite series need not converge, and where it does is 10.13's subject.

The method is the same every time. Differentiate repeatedly, evaluate at the centre, look for the pattern, write the general term, and then check it.

Take $f(x) = \sin x$ at $a = 0$. The derivatives cycle with period four:

1. $f(x) = \sin x$, so $f(0) = 0$.
2. $f'(x) = \cos x$, so $f'(0) = 1$.
3. $f''(x) = -\sin x$, so $f''(0) = 0$.
4. $f'''(x) = -\cos x$, so $f'''(0) = -1$.
5. $f^{(4)}(x) = \sin x$, and the cycle repeats.

So the values at the centre run $0, 1, 0, -1, 0, 1, \dots$ and the series is

$$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \cdots.$$

Notice what half of those zeros do: the series index and the derivative order stop agreeing. The second nonzero term is the fourth derivative's worth of work, and a general term written as though the two indices matched will not reproduce the terms above.

Three ingredients, assembled separately and then checked together.

1. The sign. Terms alternating from a positive first term need $(-1)^{n}$ with $n$ starting at $0$. Starting at $n = 1$ instead requires $(-1)^{n+1}$, and mixing the two conventions flips every sign.
2. The exponent. Odd powers are $2n+1$ and even powers are $2n$, with $n$ from $0$. For $\sin x$ the powers are $1, 3, 5, \dots$, which is $2n+1$.
3. The factorial. It matches the exponent, not the index. For $\sin x$ that is $(2n+1)!$, and writing $n!$ instead is the single most common failure here.

Together:

$$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}.$$

Now check it. $n = 0$ gives $\frac{x}{1!} = x$. $n = 1$ gives $-\frac{x^{3}}{3!}$. $n = 2$ gives $\frac{x^{5}}{5!}$. All three match the terms written out earlier, so the general term is right. This takes fifteen seconds and catches every one of the three ingredients failing, which is why it is worth doing every time rather than only when something feels wrong.

Everything above used $a = 0$, which is where most practice happens and where $(x-a)^{n}$ collapses to $x^{n}$. Away from the origin the general form reasserts itself.

For $f(x) = \ln x$ at $a = 1$: $f(1) = 0$, $f'(x) = x^{-1}$ so $f'(1) = 1$, $f''(x) = -x^{-2}$ so $f''(1) = -1$, $f'''(x) = 2x^{-3}$ so $f'''(1) = 2$, and in general $f^{(n)}(1) = (-1)^{n-1}(n-1)!$. Dividing by $n!$ leaves $\frac{(-1)^{n-1}}{n}$, so

$$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^{n}.$$

Two features worth noticing. The factorials cancel rather than vanishing, which is why a general term with no factorial in it can still be correct. And the sum starts at $n = 1$, because $f(1) = 0$ kills the constant term, which changes the sign convention from $(-1)^{n}$ to $(-1)^{n-1}$.

Writing $x^{n}$ here instead of $(x-1)^{n}$ produces a series that converges to something, just not to $\ln x$ near $1$. The centre is not notation; it is where the series is anchored.