Does anybody recognize the following power series together with a functional expression for the sum: $$ \sum_{n = 0}^{\infty} \left( \begin{array}{c} 2n \\ n \end{array} \right) x^n $$


$$ \sum_{n\geq 0}\binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}} $$ for any $x$ such that $|x|<\frac{1}{4}$ follows from the extended binomial theorem.
As an alternative, you may notice that $$ \frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos(\theta)^{2n}\,d\theta $$ and convert the previous series into something only depending on $$ \int_{0}^{\pi/2}\frac{d\theta}{1-4x\cos^2(\theta)} $$ that is simple to compute through the substitution $\theta=\arctan u$.

  • 2
    $\begingroup$ The amount of knowledge you share with this site is astonishing to me... I just wanted to say thank you. | For this to be a valid comment, I have to add a suggestion for improvement, so please post more! :) $\endgroup$ – Andrew Mar 19 '17 at 16:41
  • 1
    $\begingroup$ @Andrew: thanks so much, I am glad to help. By the way, I added an alternative derivation. $\endgroup$ – Jack D'Aurizio Mar 19 '17 at 16:42
  • $\begingroup$ Ho seguito la tua scia di briciole di pane, è stato un bel viaggio! Would you mind if I edited my derivation to the end of your post? It might be useful in the future for someone, who hasn't seen something like this before. Or should I rather post it as a separate answer? $\endgroup$ – Andrew Mar 20 '17 at 12:42
  • $\begingroup$ A separate answer is better: you may get credit for your efforts. $\endgroup$ – Jack D'Aurizio Mar 20 '17 at 13:03
  • $\begingroup$ All right, I'll do so. I asked, because earning credit for merely following your guidance didn't seem right. $\endgroup$ – Andrew Mar 20 '17 at 13:57

I wanted to elaborate on the alternate derivation Jack suggested. I'm sure there's a shorter way, but here it goes.

One can use induction to prove $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta.$$

The $n=0$ case is clear. For the inductive step note that $$\binom{2n+2}{n+1} = 4\frac{2n+1}{2n+2}\binom {2n}n,\quad \text{and} \quad \int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = \frac{2n+1}{2n+2}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta,$$ where the second is due to the following partial integration: \begin{align} \int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = 0 + \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)\sin^2(\theta)\,d\theta &= \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)(1-\cos^2(\theta))\,d\theta. \end{align}

These imply that $$\frac{1}{4^{n+1}}\binom{2n+2}{n+1} = \frac{1}{4^n}\frac{2n+1}{2n+2}\binom {2n}n = \frac{2n+1}{2n+2}\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta.$$ Meaning that the inductive proof is complete, so let's use its result. \begin{align} \sum_{n\geq 0}\left ( 4^n\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \right )x^n &= \tag{1} \frac 2\pi \int_{0}^{\pi/2} \left (\sum_{n\geq 0} 4^n \cos^{2n}(\theta) x^n\right ) \,d\theta \\&= \frac 2\pi \int_{0}^{\pi/2} \frac{1}{1-4\cos^2(\theta)x} \,d\theta \\&= \frac 2\pi \int_{\arctan0}^{\arctan \infty} \frac{1}{1-4\cos^2(\theta)x} \,d\theta \\&= \tag{2} \frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\cos^2(\arctan u)x} \,du \\&= \tag{3} \frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\frac 1{1+u^2}x} \,du \\&= \frac 2\pi \int_0^\infty \frac 1 {(1-4x)+u^2} \,du \\&= \frac 2\pi \left[ \frac{\arctan \left( \frac u {\sqrt{1-4x}} \right) }{\sqrt{1-4x}}\right]_{u=0}^{\infty} \\&= \frac 2\pi \frac{\frac \pi 2}{\sqrt{1-4x}} \\&= \frac 1 {\sqrt{1-4x}} \end{align}

Further explanation:


Recognize that this is a geometric series, which converges for $|x| < \frac 14$, since $|x| < \frac 14 \Rightarrow |4\cos^2(\theta)x| < 1.$ By analyticity, we may exchange the order of integrations. It also has a nice closed form.


Substitute $\theta = \arctan u$, use $\arctan' = \frac 1 {1+\operatorname{id_{\mathbb R}}^2}. $


For example, you can use $\tan^2(\theta) = \frac 1 {\cos^2(\theta)} - 1 $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.