Where a Series Is Valid

Each series has its own range

A series is an infinite sum, and it equals its function only for the x where that sum settles to a value. The formula list gives each standard series with that range.

$e^{x}$, sin x and cos x: every real x. ln(1 + x): $-1 < x \le 1$. $\left(1 + x\right)^{n}$, when n is not a positive integer: |x| < 1. (When n is a positive integer, the expansion stops after n + 1 terms and holds for every x.)

Why the ends differ for ln(1 + x)

At x = 1 the series is $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$, which settles to ln 2, so x = 1 is included.

At x = −1 it is $-\left(1 + \frac{1}{2} + \frac{1}{3} + \dots\right)$, the harmonic series, which grows without limit, matching ln 0, which does not exist. So x = −1 is excluded.

Whatever replaces x

To expand ln(1 + 3x), put 3x wherever the standard series has x. The range moves with it: 3x must satisfy $-1 < 3x \le 1$, so $- \frac{1}{3} < x \le \frac{1}{3}$.

A minus sign swaps the ends. ln(1 − x) needs $-1 < -x \le 1$, and multiplying by −1 reverses both inequalities: $-1 \le x < 1$.

Start the bracket with 1

The standard series is for ln(1 + x), so a bracket that starts with another number is rewritten first. $\ln\left(2 + x\right) = \ln\left(2\left(1 + \frac{x}{2}\right)\right) = \ln 2 + \ln\left(1 + \frac{x}{2}\right)$.

Now $\frac{x}{2}$ replaces x: $-1 < \frac{x}{2} \le 1$, so $-2 < x \le 2$. The binomial series works the same way: $\left(4 - x\right)^{\frac{1}{2}} = 2\left(1 - \frac{x}{4}\right)^{\frac{1}{2}}$ holds for |x| < 4.

Two series together

When two series are added or multiplied, the result holds only where both do, so take the x that lie in both ranges.

$\left(1 - 2x\right)^{-1}$ needs |−2x| < 1, so $|x| < \frac{1}{2}$, and ln(1 + x) needs $-1 < x \le 1$. The narrower range is $- \frac{1}{2} < x < \frac{1}{2}$, and the sum holds there.

$e^{2x} \ln\left(1 + x\right)$ holds for $-1 < x \le 1$: $e^{2x}$ holds for every x, so the range of ln(1 + x) decides.

The usual slips

Multiplying instead of dividing. ln(1 + 4x) needs $-1 < 4x \le 1$, so $- \frac{1}{4} < x \le \frac{1}{4}$, not $-4 < x \le 4$.

Keeping the standard range. $-1 < x \le 1$ belongs to ln(1 + x). For ln(1 + 4x) it is 4x, not x, that must lie in it.

Taking the wider range for a sum. Both series must hold, so the sum holds only on the narrower range.