Binomial Coefficient Identities via Integration

As the title indicates, this post is going to look at proving some binomial coefficient identities via integration.

The first set of identities starts with the expression \displaystyle \sum_{k=0}^n \binom{n}{k} \int_{x_1}^{x_2} x^k \, dx.

Now, evaluate this expression in two ways:

  1. Do the integration,
  2. Swap the sum and the integral, and apply the binomial theorem to the sum.

We get \displaystyle \sum_{k=0}^n \binom{n}{k} \frac{x_2^{k+1} - x_1^{k+1}}{k+1} = \frac{(x_2+1)^{n+1} - (x_1+1)^{n+1}}{n+1}.

By choosing different values of x_1 and x_2  we can obtain different binomial coefficient identities.

For example, using x_1 = 0, x_2 = 1 produces

\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{1}{k+1} = \frac{{2}^{n+1} - 1}{n+1}.

(For another quick proof, see my answer to “How can I compute \sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} ?”)

With x_1 = -1, x_2 = 0, we get

\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k+1} = \frac{1}{n+1}.

(See also the answers to “How to prove \sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}?”  André Nicolas gives a slight variation on this proof, Arturo Magidin proves the identity by manipulating properties of the binomial coefficients, and I give a probabilistic/combinatorial argument.)

If we take x_1 = -1, x_2 = 1, we obtain

\displaystyle \sum_k \binom{n}{2k} \frac{1}{2k+1} = \frac{1}{2} \sum_{j=0}^n \binom{n}{j} \frac{1 - (-1)^{j+1}}{j+1} = \frac{2^n}{n+1}.

(These three examples are all from Chapter 3 of Charalambides’s Enumerative Combinatorics [1].  He gives several other interesting binomial coefficient identities and discusses some different proof techniques as well.)

Another method for proving binomial coefficient identities via integration arises from the representation of the binomial coefficient as a beta function.  The beta function B(x,y) is defined as \displaystyle \int_0^1 t^{x-1} (1-t)^{y-1} \, dt.  The beta function also has a representation in terms of the gamma function, \displaystyle B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.  Thus, for integer n and k, we can write \displaystyle \binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1) \Gamma(n-k+1)} = \frac{1}{(n+1) B(k+1,n-k+1)}.  (Actually, this can be used to define the binomial coefficient for non-integer values of n and k.  But we’ll stick with integer values of n and k for this post.)

The beta function representation of the binomial coefficient can sometimes lead to binomial coefficient identities, too.  A beautiful example of this was recently given on math.SE by Eric Naslund.  Suppose we want to find \displaystyle \sum_{k=0}^{\infty} \frac{2^k}{(k+1) \binom{2k}{k}}.  Using the beta function representation, we can rewrite this as

\displaystyle \sum_{k=0}^{\infty} 2^k \int_0^1 x^k (1-x)^k \, dx = \int_0^1 \left(\sum_{k=0}^{\infty} 2^k x^k (1-x)^k \right) \, dx = \int_0^1 \frac{1}{1-2x(1-x)} \, dx

\displaystyle =\int_0^1 \frac{1}{x^2+(1-x)^2} \, dx .

Now, substitute u = 1/x, followed by v = u-1, and we obtain \displaystyle \int_1^{\infty} \frac{1}{1 + (u-1)^2} \, du = \int_0^{\infty} \frac{1}{1 + v^2} \, dv = \left.\arctan v \right|_0^{\infty} = \frac{\pi}{2}.

Thus \displaystyle \sum_{k=0}^{\infty} \frac{2^k}{(k+1) \binom{n}{k}} = \frac{\pi}{2}.

For more identities involving sums of reciprocals of the central binomial coefficients, including some generating functions, see Sprugnoli’s paper [2].  (However, Sprugnoli explicitly avoids the use of integration.)


  1. Charalambos A. Charalambides,  Enumerative Combinatorics, Chapman & Hall/CRC, 2002.
  2. Renzo Sprugnoli, “Sums of reciprocals of the central binomial coefficients,” Integers 6 (2006), Article A27.
This entry was posted in binomial coefficients, calculus. Bookmark the permalink.

2 Responses to Binomial Coefficient Identities via Integration

  1. Pingback: A Probabilistic Proof of a Binomial Coefficient Identity | A Narrow Margin

  2. Pingback: Shifts, Finite Differences, and Binomial Transforms | A Narrow Margin

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s