Binomial Series – Definition, General Form, and Examples

The binomial series is one of the most important infinite series you’ll encounter in Calculus. You’ll be working with binomial series often, so it helps if you know its expansion by heart!

The binomial series is an infinite series that results in expanding a binomial by a given power.  In fact, it is a special type of a Maclaurin series for functions, $\boldsymbol{f(x) = (1 + x)^m}$, using a special series expansion formula.

In this article, we’ll focus on expanding $(1 + x)^m$, so it’s helpful to take a refresher on the binomial theorem. Since the binomial series is a special type of a Maclaurin series, so keep your notes handy or head over to this link in case you need a refresher.

For now, let’s go ahead and understand what makes a binomial series special!

What is a binomial series?

The binomial series is simply the Maclaurin series of the function, $\boldsymbol{(1 +x)^m}$, where $m$ is any real number. This means that the binomial series shows the sum of terms resulting from expanding $(1 +x)^m$ from powers from $0$ to $m$.

In our algebra classes, we’ve learned how to expand a binomial, $(a + b)^m$, using the binomial theorem: $(a +b)^m = \sum_{n = 0}^{\infty}\binom{m}{k}a^{m – k}b^k$.

\begin{aligned}(a+b)^1 &= a+ b\\(a+b)^2 &= a^2 +2ab+b^2\\(a +b)^3&=a^3 + 3a^2b+3ab^2 +b^3\\&.\\&.\\&.\end{aligned}

From this alone, we can see that it’s going to be challenging if we continue using the binomial theorem to expand binomial expressions especially when $m$ is significantly large or small. This is when the binomial series enters: we can expand $\boldsymbol{(1 + x)^m}$ regardless of $\boldsymbol{m}$’s value.

Binomial series formula

We begin deriving the binomial series’ formula by establishing the Maclaurin series expansion of $(1 + x)^m$. Through the Maclaurin expansion, we can approximate $f(x)$ as shown below.

\begin{aligned}f(x) \approx f(0) + f^{\prime}(0)x + \dfrac{ f^{\prime\prime}(0)x}{2!}+ \dfrac{ f^{\prime\prime\prime}(0)x}{3!} + …\end{aligned}

Apply this expansion technique and we’ll show you the work here as a guide.

\begin{aligned}f(x) &= (1 + x)^m\end{aligned}\begin{aligned}f(0) &= 1\end{aligned}
\begin{aligned}f^{\prime}(x) &= m(1 + x)^{m -1}\end{aligned}\begin{aligned}f^{\prime}(0) &= m\end{aligned}
\begin{aligned}f^{\prime\prime}(x) &= m(m -1)(1 + x)^{m -2}\end{aligned}\begin{aligned}f^{\prime\prime}(0) &= m(m -1)\end{aligned}
\begin{aligned}f^{\prime\prime\prime }(x) &= m(m -1)(m -2)(1 + x)^{m -3}\end{aligned}\begin{aligned}f^{\prime\prime\prime }(0) &= m(m -1)(m -2)\end{aligned}
\begin{aligned}f^{(n)}(x) &= m(m -1)(m -2)…(m – n +1)(1 + x)^{m -n}\end{aligned}\begin{aligned}f^{(n)}(0) &= m(m -1)(m -2)… (m -n+1)\end{aligned}

This means that the Maclaurin series for the function, $f(x) =(1 + x)^m$, is as shown below.

\begin{aligned}\sum_{n =0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n = \sum_{n =0}^{\infty} \dfrac{m(m -1)(m- 2)…(m -n + 1)}{n!}x^n\end{aligned}

Let’s observe the ratio of the $(n +1)$th and $n$th terms:

\begin{aligned}\left|\dfrac{a_{n + 1}}{a_n}\right| &= \dfrac{|m(m -1)…(m -n)||x|^{n +1}}{(n +1)!}\cdot \dfrac{n!}{|m(m-1) …(m – n+ 1)||x|^n}\\&= \dfrac{|m -n|}{n + 1}|x|\end{aligned}

This means that $\lim_{n \rightarrow \infty}\left|\dfrac{a_{n+1}}{a_n}\right| \rightarrow |x|$.Through the ratio test, we can see that the binomial series will only be convergent when $\boldsymbol{|x| <1}$.


This means that when $m$ is any real number and $|x| <1$, we can expand $(1+x)^m$ as shown below:

\begin{aligned}\boldsymbol{\sum_{n =0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n } = \boldsymbol{\sum_{n =0}^{\infty}\binom{m}{n}x^n}=\boldsymbol{1+ mx + \dfrac{m(m -1)}{2!}x^2 + \dfrac{m(m -1)(m -2)}{3!}x^3+…}\end{aligned}

How to apply the binomial series expansion?

Here’s a quick guideline to remember when expanding a binomial series:

  • Express the function as a power of $(1 +x)$: $(1 +x)^m$.
  • Expand the formula up to the terms that are needed. When the number of terms is not given, use up to $x^3$.
  • Make sure to set the conditions for $x$ in your answer: for the binomial series to work, $-1<x<1$ or $|x|<1$.

We can apply the binomial series formula to expand the function, $f(x) = \sqrt{1 +x}$, as far as the term in $x^4$. First, let’s write $\sqrt{1 +x}$ as a power of $(1 +x)$.

\begin{aligned}f(x) &= \sqrt{1 + x} &= (1 + x)^{\frac{1}{2}}\end{aligned}

We can go ahead and apply the binomial series formula and use $m = \dfrac{1}{2}$. We only need the series to expand up to $x^4$.

\begin{aligned}f(x) &\approx 1+ mx + \dfrac{m(m -1)}{2!}x^2 + \dfrac{m(m -1)(m -2)}{3!}x^3+\dfrac{m(m -1)(m -2)(m -3)}{4!}x^4+…\\&=1 + \dfrac{1}{2}x + \dfrac{\frac{1}{2}\left(\frac{1}{2}-1 \right )}{2!} x^2 + \dfrac{\frac{1}{2}\left(\frac{1}{2}-1 \right )\left(\frac{1}{2} -2\right )}{3!} x^3 + \dfrac{\frac{1}{2}\left(\frac{1}{2}-1 \right )\left(\frac{1}{2} -2\right )\left(\frac{1}{2}-3 \right )}{4!} x^4 +…\end{aligned}

Simplify each terms’ numerator then we now have $\sqrt{1+x}$’s binomial series expansion!

\begin{aligned}f(x) &=1 + \dfrac{x}{2}+ \dfrac{-\frac{1}{4}x^2}{2}+ \dfrac{\frac{3}{8}x^3}{6} + \dfrac{-\frac{15}{16}x^4}{24}+…\\&=1 + \dfrac{x}{2}-\dfrac{x^2}{8} + \dfrac{x^3}{16}-\dfrac{5x^4}{128}+ …\end{aligned}

This means that the Maclaurin expansion of $\sqrt{1 + x}$ is equal to $1 + \dfrac{x}{2}-\dfrac{x^2}{8} + \dfrac{x^3}{16}-\dfrac{5x^4}{128}+…$ as long as $|x| < 1$.

We’ve prepared more functions for you to expand in the next section, so head over to the problems below when you’re ready!

Example 1

Expand the function, $f(x) = (3 – x)^{-3}$, as a binomial series up to $x^3$.


Let’s rewrite the binomial so that it has $1$ as the coefficient and addition as the operation between $1$ and the variable.

\begin{aligned}(3 – x)^{-3} &= \left[3\left(1 – \dfrac{x}{3}\right)\right]^{-3}\\&= \dfrac{1}{27}\left(1 – \dfrac{x}{3} \right )^{-3}\\&= \dfrac{1}{27}\left[1- \left(-\dfrac{x}{3} \right ) \right ]^{-3}\end{aligned}

We now have the binomial form that we need to apply the binomial series formula. Hence, we have the following expansion:

\begin{aligned}f(x) &=1+ mx + \dfrac{m(m -1)}{2!}x^2 + \dfrac{m(m -1)(m -2)}{3!}+…\\m&= -3\\x&= – \dfrac{x}{3}\\\\(3 – x)^{-3}&=\dfrac{1}{27}\left[1+\left(-\dfrac{x}{3} \right ) \right ]^{-3} \\&= \dfrac{1}{27}\left[1 + (-3)\left(-\dfrac{x}{3} \right ) + \dfrac{-3(-3 -1)}{2!}\left(-\dfrac{x}{3} \right )^2+ \dfrac{-3(-3 -1)(-3 -2)}{3!}\left(-\dfrac{x}{3} \right )^3+… \right ]\end{aligned}

Simplify each term to have the binomial series shown below.

\begin{aligned}(3 – x)^{-3}&= \dfrac{1}{27}\left[1 + x +\dfrac{2}{3}x^2+ \dfrac{10}{27}x^3+… \right ]\end{aligned}

Since we’re working with $-\dfrac{x}{3}$ instead of $x$, the restrictions for $x$ will change as well.

\begin{aligned}-1&< -\dfrac{x}{3} <1\\-3(-1)&>x >-3(1)\\-3 &< x<3\end{aligned}

This means that $(3 – x)^{-3} = \dfrac{1}{27}\left[1 + x +\dfrac{2}{3}x^2+ \dfrac{10}{27}x^3+… \right]$ provided that $|x| < 3$.

Example 2

Approximate $\sqrt{1.3}$ correct to three decimal places.


Express $\sqrt{1.3}$ as $(1 + 0.3)^{\frac{1}{2}}$ then apply the binomial series formula to approximate its value. Expand the terms as far as $x^3$.

\begin{aligned}f(x)&\approx 1+ mx + \dfrac{m(m -1)}{2!}x^2 + \dfrac{m(m -1)(m -2)}{3!}+\dfrac{m(m -1)(m -2)(m -3)}{4!}+…\\m &= \dfrac{1}{2}\\x&= 0.3\\\\(1 + 0.3)^{\frac{1}{2}} &=1 + \dfrac{1}{2}(0.3) + \dfrac{\frac{1}{2}\left(\frac{1}{2}-1 \right )}{2!} (0.3)^2 + \dfrac{\frac{1}{2}\left(\frac{1}{2}-1 \right )\left(\frac{1}{2} -2\right )}{3!} (0.3)^3 +…\\&= 1 + 0.15 – \dfrac{1}{8}(0.09) + \dfrac{1}{16}(0.027)+…\\&= 1 +0.15 – 0.01125+0.0016875 + …\\&\approx 1.1404\end{aligned}

This means that through the binomial series, we can now approximate the values of expressions such as $\sqrt{1.03}$. In fact, we have $\sqrt{1.3} \approx 1.140$ correct to three decimal places.

Practice Questions

1. Expand the function, $f(x) = \dfrac{1}{1 +x}$, as a binomial series up to $x^3$.
2. Expand the function, $f(x) = \sqrt[3]{1 +x}$, as a binomial series up to $x^4$.
3. Expand the function, $f(x) = (2 – x)^{-3}$, as a binomial series up to $x^3$.
4. Expand the function, $f(x) = \sqrt{1 – 2x}$, as a binomial series up to $x^3$.
5. Approximate $\sqrt{1.02}$ correct to three decimal places.

Answer Key

1. $f(x) = \dfrac{1}{1 + x} = 1 – x + x^2 – x^3 + …$ provided that $|x| < 1$
2. $f(x) = \sqrt[3]{1 + x} = 1 + \dfrac{x}{3} – \dfrac{x^2}{9} + \dfrac{5x^3}{81} – \dfrac{10x^4}{243}+ …$ as long as $|x|<1$
3. $f(x) = \dfrac{1}{2}\left[1 + \left(-\dfrac{x}{2}\right)\right]^{-3} = \dfrac{1}{8}\left[1 + \dfrac{3x}{2} + \dfrac{3x^2}{2} + \dfrac{5x^3}{4}+ …\right]$ provided that $|x| < 2$.
4. $f(x) = \sqrt{1 – 2x} = 1 – x – \dfrac{x^2}{2} – \dfrac{x^3}{2} – …$ as long as $|x|<1$ as long as $|x| < 3$
5. $\sqrt{1.02} \approx 1.00995$