In class, I defined the binomial coefficient using an integral:

$$\binom{n}{k} = \displaystyle \int_0^{2\pi}\dfrac{dt}{2\pi} e^{-ikt}(1+e^{it})^n.$$

I succeeded in demonstrating many standard properties of the binomial coefficient directly using integration: Pascal's identity, Vandermonde identity, Hockey stick. But I could not show that $$\sum_{k=0}^n \binom{n}{k}=2^n.$$

It turns out I have to show the following:


I do not know how to perform this integration! I need help. It is better if the solution did not involve contour integration.


3 Answers 3



Notice that inside the integral we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}$$ which is equal to $D_{\frac n2}$ where $D_n$ denotes the Dirichlet kernel.

Hence using it's properties we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}=D_{\frac n2}=1+2\sum_{r=1}^{\frac n2} \cos (rt) $$

Hope you can continue further

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    $\begingroup$ Thanks for the tip. But this was an intermediate step in the proof. Your idea reduces the computation to the integral $\displaystyle \int_0^{2\pi} \cos(rt) \cos^n \frac{t}2 dt/2\pi$, and I didn't see how to do this then. I will try it again. $\endgroup$ May 21, 2018 at 5:21
  • $\begingroup$ @Isomorphism You might have to take cases of whether 'n' is even or odd. For n as odd will be a very simple case and will be solved pretty easily. But the case of n as even will be a bit tedious one $\endgroup$ May 21, 2018 at 6:06

There is a mistake in my presentation that eludes me, but the essence is the following.

Using $$\sum_{k=1}^{\infty} p^{k} \, \sin(k \, x) = \frac{p \, \sin(x)}{1 - 2 \, p \, \cos(x) + p^{2}}$$ then let $S_{n}$ be the desired summation to be evaluated to obtain: \begin{align} S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \\ &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin\left(\frac{t}{2}\right)} \, dt \\ &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin(t)} \, dt \end{align} Now, \begin{align} \sum_{n=0}^{\infty} S_{n} &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=0}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \sin\left(\frac{(n+1) \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\ &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \sin\left(\frac{n \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\ &= \frac{1}{2 \pi} \, \int_{0}^{2 \pi} \frac{2 \cos\left(\frac{t}{2}\right) \sin\left(\frac{t}{2}\right)}{1 - 4 \cos^{2}\left(\frac{t}{2}\right) + 4 \cos^{2}\left(\frac{t}{2}\right)} \, \frac{dt}{\sin(t)} \\ &= \frac{1}{2\pi} \, \int_{0}^{2 \pi} dt = 1 \\ &= - (-1) = - \frac{1}{1 - 2} = - \sum_{n=0}^{\infty} 2^{n}. \end{align}

This yields $$\sum_{n=0}^{\infty} \binom{n}{k} = - 2^{n}.$$

As stated, the mistake that eludes me is the sign error.

  • $\begingroup$ I don't think the second last step is right. The infinite geometric sum can be applied only when the common ratio of the GP lies between $-1$ and $1$ but as per your text it seems that you have taken 2 as the common ratio which I don't think would be proper to continue either $\endgroup$ May 31, 2018 at 18:39

Using mathematical induction. Base case $n=0$, ok!
Assume that is true for an arbitrary natural number $n$. Since $$ \sum_{k=0}^{n+1} \binom{n+1}{k} = \sum_{k=0}^n \binom{n+1}{k} + \binom{n+1}{n+1} $$ Using your definition, $\binom{n+1}{n+1}$ equals to $1$, then using Pascal's identity $$ \binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1} \text{ for all } 1\leqslant k\leqslant n $$ You get the following : $$ \sum_{k=0}^{n+1} \binom{n+1}{k} = 2^n + 2^n - 1 +1 = 2^{n+1}$$

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    $\begingroup$ I know this. I want to prove the statement via the integral. In other words, I want the proof of the integral identity. $\endgroup$ May 20, 2018 at 22:05
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    $\begingroup$ But you proved Pascal's identity using integral representation... this works fine. $\endgroup$
    – Célestin
    May 20, 2018 at 22:07
  • $\begingroup$ Otherwise, you can perform some contour integration... but you won't... $\endgroup$
    – Célestin
    May 20, 2018 at 22:10
  • $\begingroup$ I don't mind seeing a contour integral proof. I just prefer an elementary integration proof. $\endgroup$ May 20, 2018 at 22:13

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