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.

$e^x$eˣ, sin x and cos x: every real x. ln(1 + x): $-1<x\le 1$−1 < x ≤ 1. ${\left(1+x\right)}^n$(1 + x)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$1 − 1/2 + 1/3 − 1/4 + …, which settles to ln 2, so x = 1 is included.

At x = −1 it is $-\left(1+\frac{1}{2}+\frac{1}{3}+\dots \right)$−(1 + 1/2 + 1/3 + …), 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 $-1<3x\le 1$−1 < 3x ≤ 1, so $-\frac{1}{3}<x\le \frac{1}{3}$−1/3 < x ≤ 1/3.

A minus sign swaps the ends. ln(1 − x) needs $-1<-x\le 1$−1 < −x ≤ 1, and multiplying by −1 reverses both inequalities: $-1\le x<1$−1 ≤ 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. $\ln \left(2+x\right)=\ln \left(2\left(1+\frac{x}{2}\right)\right)=\ln 2+\ln \left(1+\frac{x}{2}\right)$ln(2 + x) = ln(2(1 + x/2)) = ln 2 + ln(1 + x/2).

Now $\frac{x}{2}$x/2 replaces x: $-1<\frac{x}{2}\le 1$−1 < x/2 ≤ 1, so $-2<x\le 2$−2 < x ≤ 2. The binomial series works the same way: ${\left(4-x\right)}^{\frac{1}{2}}=2{\left(1-\frac{x}{4}\right)}^{\frac{1}{2}}$(4 − x)^½ = 2(1 − x/4)^½ 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.

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

$e^{2x}\ln \left(1+x\right)$e²ˣ ln(1 + x) holds for $-1<x\le 1$−1 < x ≤ 1: $e^{2x}$e²ˣ holds for every x, so the range of ln(1 + x) decides.

The usual slips

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

Keeping the standard range. $-1<x\le 1$−1 < x ≤ 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.