Binomial Series: Formula & Examples

The binomial series extends the binomial theorem to any real power k: $(1 + x)^k = 1 + kx + \frac{k(k - 1)}{2!}x^2 + \frac{k(k - 1)(k - 2)}{3!}x^3 + \cdots$ . When k is a whole number, the series stops. Otherwise it never ends, and it converges for |x| < 1.

Before this lesson: Binomial Theorem, Maclaurin Series and Radius and Interval of Convergence

From the binomial theorem to the binomial series

The binomial theorem expands $(1 + x)^4$ into five terms with the coefficients 1, 4, 6, 4, 1. Those coefficients can also be written as $\frac{4 \cdot 3}{2!} = 6$ and so on. That form makes sense for any k, not just whole numbers.

When k is a fraction or a negative number, the factors k, k − 1, k − 2, … never reach 0, so the terms never stop. The result is the Maclaurin series of $(1 + x)^k$ , and it works only near 0.

The binomial series for √(1 + x) tracks the curve between x = −1 and x = 1 and breaks away past x = 1.

The binomial series formula

(1 + x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n, \quad |x| \lt 1

\binom{k}{n} = \dfrac{k(k - 1)(k - 2)\cdots(k - n + 1)}{n!}, \quad \binom{k}{0} = 1

The top of $\binom{k}{n}$ has n factors, starting at k and going down by 1 each time.
If k is a whole number, the series stops after the $x^k$ term; this is the binomial theorem.
Otherwise the radius of convergence is 1. Whether the endpoints x = ±1 are included depends on k.
For $(a + x)^k$ , factor out a first: $(a + x)^k = a^k\left(1 + \frac{x}{a}\right)^k$ , valid for |x| < |a|.

How to use the binomial series

Write the expression as $(1 + u)^k$ , factoring out a constant if needed.
Find the coefficients $\binom{k}{0}, \binom{k}{1}, \binom{k}{2}, \binom{k}{3}, \ldots$ one at a time.
Multiply each by the matching power of u and simplify.
State where it is valid: |u| < 1.

Binomial series examples

Example 1: The series for a square root

Find the first four terms of the binomial series for $\sqrt{1 + x} = (1 + x)^{1/2}$ .

\binom{1/2}{1} = \dfrac{1}{2}

One factor: k = 1/2.

\binom{1/2}{2} = \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!} = \dfrac{-1/4}{2} = -\dfrac{1}{8}

Two factors: 1/2 and 1/2 − 1 = −1/2. Divide by 2! = 2.

\binom{1/2}{3} = \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!} = \dfrac{3/8}{6} = \dfrac{1}{16}

Three factors; multiply the fractions, then divide by 3! = 6.

\sqrt{1 + x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 - \cdots

Put the coefficients on the powers of x.

Answer: $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots$ for |x| < 1 (and at both endpoints too).

Example 2: A negative power

Find the first four terms of the series for $\dfrac{1}{(1 + x)^3}$ .

\dfrac{1}{(1 + x)^3} = (1 + x)^{-3}

Rewrite with a negative exponent (see zero and negative exponents).

\binom{-3}{1} = -3, \quad \binom{-3}{2} = \dfrac{(-3)(-4)}{2} = 6

Use the formula with k = −3.

\binom{-3}{3} = \dfrac{(-3)(-4)(-5)}{6} = \dfrac{-60}{6} = -10

Three factors, divided by 3! = 6.

(1 + x)^{-3} = 1 - 3x + 6x^2 - 10x^3 + \cdots

Put the coefficients on the powers of x.

Answer: $\frac{1}{(1 + x)^3} = 1 - 3x + 6x^2 - 10x^3 + \cdots$ for |x| < 1.

Example 3: Estimate a cube root

Use four terms of the binomial series to estimate $\sqrt[3]{1.06}$ .

\sqrt[3]{1.06} = (1 + 0.06)^{1/3}

k = 1/3 and x = 0.06, which is inside |x| < 1.

\binom{1/3}{1} = \dfrac{1}{3}, \quad \binom{1/3}{2} = \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2} = -\dfrac{1}{9}

Find the first coefficients.

\binom{1/3}{3} = \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{6} = \dfrac{10/27}{6} = \dfrac{5}{81}

Three factors, divided by 3! = 6.

1 + \dfrac{0.06}{3} - \dfrac{0.0036}{9} + \dfrac{5(0.000216)}{81}

0.06² = 0.0036 and 0.06³ = 0.000216.

1 + 0.02 - 0.0004 + 0.0000133 \approx 1.0196133

Divide, then combine.

Answer: $\sqrt[3]{1.06} \approx 1.019613$ . (The true value is 1.019613 to six decimal places.)

Example 4: Building an arcsine series

Find the series for $\dfrac{1}{\sqrt{1 - x^2}}$ , then integrate it to get a series for arcsin x.

\dfrac{1}{\sqrt{1 - x^2}} = (1 + u)^{-1/2}, \quad u = -x^2

Rewrite in the form (1 + u)ᵏ.

\binom{-1/2}{1} = -\dfrac{1}{2}, \; \binom{-1/2}{2} = \dfrac{3}{8}, \; \binom{-1/2}{3} = -\dfrac{5}{16}

For example, (−1/2)(−3/2)/2 = 3/8 and (−1/2)(−3/2)(−5/2)/6 = −5/16.

(1 - x^2)^{-1/2} = 1 + \dfrac{1}{2}x^2 + \dfrac{3}{8}x^4 + \dfrac{5}{16}x^6 + \cdots

Put u = −x² into each term; the minus signs cancel in pairs.

\arcsin x = x + \dfrac{x^3}{6} + \dfrac{3x^5}{40} + \dfrac{5x^7}{112} + \cdots

Integrate term by term (see integrating power series); C = arcsin 0 = 0.

Answer: $\arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots$ for |x| < 1.

Example 5: Time dilation (word problem)

In relativity, a clock moving at speed v runs slow by the factor $\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-1/2}$ , where c is the speed of light. Use three terms of the binomial series to find γ when v = 0.1c.

\gamma = (1 + u)^{-1/2}, \quad u = -\dfrac{v^2}{c^2} = -0.01

With v = 0.1c, v²/c² = 0.01.

\gamma \approx 1 - \dfrac{1}{2}u + \dfrac{3}{8}u^2

Use the coefficients from Example 4.

1 - \dfrac{1}{2}(-0.01) + \dfrac{3}{8}(0.0001) = 1 + 0.005 + 0.0000375

(−0.01)² = 0.0001.

1 + 0.005 + 0.0000375 = 1.0050375

Combine.

Answer: γ ≈ 1.00504, so the moving clock runs about 0.5% slow. (The true value is 1.0050378.)

Common mistakes

Mistake	Example of the error	Fix
Using ordinary binomial coefficients	Treating $\binom{1/2}{2}$ as “1/2 choose 2” from Pascal’s triangle	Use the formula: $\frac{(1/2)(-1/2)}{2!} = -\frac{1}{8}$ .
Forgetting n! in the bottom	Writing $\binom{-3}{3} = (-3)(-4)(-5) = -60$	Divide by 3! = 6 to get −10.
Not factoring out the constant	Expanding $\sqrt{4 + x}$ as $(1 + (3 + x))^{1/2}$	Write $2\left(1 + \frac{x}{4}\right)^{1/2}$ so the inside is small.
Using it outside \|x\| < 1	Putting x = 3 into the series for $\sqrt{1 + x}$ to find 2	The series diverges there. Factor so the inside is less than 1.
Missing the sign of u	For $(1 - x^2)^{-1/2}$ , using u = x²	Here u = −x², so each power of u brings its own sign.
Thinking the series stops for k = 1/2	Stopping after the x term	Only whole-number k stops; for other k the series goes on forever.

Related methods

The geometric series is a binomial series

With k = −1 and x replaced by −x, the binomial series gives $(1 - x)^{-1} = 1 + x + x^2 + \cdots$ , the geometric series from power series. With k = −2 it gives $\frac{1}{(1 - x)^2} = \sum (n + 1)x^n$ .

Quick approximations

The first two terms, $(1 + x)^k \approx 1 + kx$ , are the linear approximation of $(1 + x)^k$ at x = 0. Physics and engineering use this shortcut constantly for small x.

Practice problems

Find the first four terms of the binomial series for $(1 + x)^{1/3}$ .
Find the first four terms of the series for $\dfrac{1}{(1 + x)^2}$ .
Find the first four terms of the series for $\sqrt{1 + 2x}$ , and say where it is valid.
Find the first four terms of the series for $\sqrt{4 + x}$ .
Use three terms to estimate $\sqrt{1.04}$ .
What is the radius of convergence of the series for $\left(1 + \frac{x}{3}\right)^{-1/2}$ ?
Use the binomial series to expand $(1 + x)^4$ . What happens?
Use the binomial series with k = −2 to find the series for $\dfrac{1}{(1 - x)^2}$ .
Find the coefficient of $x^3$ in the series for $(1 + x)^{3/2}$ .
A cube-shaped box has a volume of 1030 cm³. Its edge is $\sqrt[3]{1030} = 10(1 + 0.03)^{1/3}$ cm. Use three terms to estimate the edge.

Show answers

\binom{1/3}{1} = \dfrac{1}{3}, \; \binom{1/3}{2} = -\dfrac{1}{9}, \; \binom{1/3}{3} = \dfrac{5}{81}

From Example 3.

Answer: $1 + \dfrac{x}{3} - \dfrac{x^2}{9} + \dfrac{5x^3}{81} - \cdots$

\binom{-2}{1} = -2, \; \binom{-2}{2} = \dfrac{(-2)(-3)}{2} = 3

Use k = −2.

\binom{-2}{3} = \dfrac{(-2)(-3)(-4)}{6} = -4

Three factors, divided by 3! = 6.

Answer: $1 - 2x + 3x^2 - 4x^3 + \cdots$ for |x| < 1

(1 + 2x)^{1/2} = 1 + \dfrac{1}{2}(2x) - \dfrac{1}{8}(2x)^2 + \dfrac{1}{16}(2x)^3

Use the coefficients from Example 1 with u = 2x.

1 + x - \dfrac{4x^2}{8} + \dfrac{8x^3}{16} = 1 + x - \dfrac{x^2}{2} + \dfrac{x^3}{2}

(2x)² = 4x² and (2x)³ = 8x³.

|2x| \lt 1 \quad \Leftrightarrow \quad |x| \lt \dfrac{1}{2}

The series needs |u| < 1.

Answer: $1 + x - \dfrac{x^2}{2} + \dfrac{x^3}{2} - \cdots$ for |x| < 1/2

\sqrt{4 + x} = 2\left(1 + \dfrac{x}{4}\right)^{1/2}

Factor out 4; √4 = 2.

2\left[1 + \dfrac{1}{2} \cdot \dfrac{x}{4} - \dfrac{1}{8} \cdot \dfrac{x^2}{16} + \dfrac{1}{16} \cdot \dfrac{x^3}{64}\right]

Use u = x/4.

2 + \dfrac{x}{4} - \dfrac{x^2}{64} + \dfrac{x^3}{512}

Multiply each term by 2.

Answer: $2 + \dfrac{x}{4} - \dfrac{x^2}{64} + \dfrac{x^3}{512} - \cdots$ for |x| < 4

\sqrt{1.04} \approx 1 + \dfrac{0.04}{2} - \dfrac{(0.04)^2}{8}

x = 0.04.

1 + 0.02 - \dfrac{0.0016}{8} = 1 + 0.02 - 0.0002 = 1.0198

Divide, then combine.

Answer: about 1.0198 (true value 1.019804)

\left|\dfrac{x}{3}\right| \lt 1

The series in u = x/3 needs |u| < 1.

|x| \lt 3

Multiply by 3.

Answer: R = 3

\binom{4}{1} = 4, \; \binom{4}{2} = 6, \; \binom{4}{3} = 4, \; \binom{4}{4} = 1

For example, (4)(3)(2)/3! = 4.

\binom{4}{5} = \dfrac{4 \cdot 3 \cdot 2 \cdot 1 \cdot 0}{5!} = 0

From n = 5 on, a factor of 0 appears.

Answer: $1 + 4x + 6x^2 + 4x^3 + x^4$ ; the series stops, matching the binomial theorem

(1 + u)^{-2} = 1 - 2u + 3u^2 - 4u^3 + \cdots

From #2, with u in place of x.

(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \cdots

Put in u = −x; every sign becomes +.

Answer: $\sum_{n=0}^{\infty} (n + 1)x^n$ for |x| < 1

\binom{3/2}{3} = \dfrac{\frac{3}{2} \cdot \frac{1}{2} \cdot \left(-\frac{1}{2}\right)}{3!}

Three factors: 3/2, 1/2 and −1/2.

\dfrac{-3/8}{6} = -\dfrac{3}{48} = -\dfrac{1}{16}

Multiply, then divide by 6.

Answer: −1/16

10.

10(1 + 0.03)^{1/3} \approx 10\left[1 + \dfrac{0.03}{3} - \dfrac{(0.03)^2}{9}\right]

Use the coefficients 1, 1/3 and −1/9 from Example 3.

10\left[1 + 0.01 - 0.0001\right] = 10(1.0099) = 10.099

0.03² = 0.0009 and 0.0009/9 = 0.0001.

Answer: about 10.099 cm (true value 10.09902)

Binomial series FAQ

What is the binomial series?

It is the Maclaurin series of $(1 + x)^k$ for any real number k: $\sum \binom{k}{n} x^n$ , where $\binom{k}{n} = \frac{k(k - 1)\cdots(k - n + 1)}{n!}$ .

When does the binomial series converge?

If k is a whole number, it is a finite sum and works for every x. Otherwise it converges for |x| < 1; the endpoints depend on k.

What is the difference between the binomial theorem and the binomial series?

The binomial theorem expands $(a + b)^n$ for a whole number n and always stops. The binomial series handles any real power, so it usually never stops.

How do you find a binomial coefficient for a fraction?

Multiply n factors starting at k and going down by 1, then divide by n!. For example, $\binom{1/2}{2} = \frac{(1/2)(-1/2)}{2} = -\frac{1}{8}$ .

Summary

$(1 + x)^k = \sum \binom{k}{n} x^n$ for |x| < 1, with $\binom{k}{n} = \frac{k(k-1)\cdots(k-n+1)}{n!}$ .
Whole-number k gives a finite sum: the binomial theorem.
Factor out constants so the expression is $(1 + u)^k$ with |u| < 1.
Use it for roots, negative powers and quick estimates like (1 + x)ᵏ ≈ 1 + kx.
Integrating a binomial series can give series for functions like arcsin x.

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