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

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$$ \sum_{n\geq 0}\binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}} $$ 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 $\theta=\arctan u$.

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    $\begingroup$ 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
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    $\begingroup$ @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
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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:

(1)

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.

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

(3)

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

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