Working on a problem I have encountered an interesting identity:

$$ \sum_{k=0}^\infty \left(\frac{x}{2}\right)^{n+2k}\binom{n+2k}{k} =\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n, $$ where $n$ is a non-negative integer number and $x$ is a real number with absolute value less than 1 (probably a similar expression is valid for arbitrary complex numbers $|z|<1$).

Is there any simple proof of this identity?

  • $\begingroup$ How/where did you meet this identity? Is the LHS just the Taylor series for the RHS? $\endgroup$ – Daniel Robert-Nicoud Dec 14 '17 at 23:38
  • $\begingroup$ Anything to do with a random walk? Why did you call this interesting, btw? $\endgroup$ – Mathemagical Dec 14 '17 at 23:44
  • $\begingroup$ Reminiscent of the explicit form for Chebyshev polynomials of the first kind, hence they idea of enforcing the substitution $x=\sin\theta$ and applying the residue theorem looks like a promising one ;) $\endgroup$ – Jack D'Aurizio Dec 14 '17 at 23:56
  • 1
    $\begingroup$ Or you may prove that both sides are the terms of an Appell / Fibonacci sequence of polynomials. $\endgroup$ – Jack D'Aurizio Dec 14 '17 at 23:57
  • 1
    $\begingroup$ @tired This is in regard to the hitting time of a random walk (more particularly, the probability generating function of the hitting time of the walk at the barrier x=1. Please see the equations 15,16, and 17 in the linked document. galton.uchicago.edu/~lalley/Courses/312/RW.pdf The details of the computation are not listed here, but they are shown in greater detail in Michael Steele's book, Stochastic Calculus and Financial applications (pages 8 and 9). $\endgroup$ – Mathemagical Dec 16 '17 at 15:25


$$\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz$$ we get (integration contour is the unit cicrle)

$$ 2\pi iS_n=\oint dz \sum_{k=0}^{\infty}\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\oint dz \frac{(1+z)^n x^n}{z2^n}\sum_{k=0}^{\infty}\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\ 4\frac{x^n}{2^n}\oint dz \underbrace{\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)} $$

for $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\frac2{x^2}-\frac{2\sqrt{1-x^2}}{x^2}-1$ , so

$$ S_n=4\frac{x^n}{2^n}\text{res}(f(z),z=z_0(x))=4\frac{x^n}{2^n}\left[ \frac{1}{4 \sqrt{1-x^2}}\left(2\frac{1-\sqrt{1-x^2}}{ x^2}\right)^n\right] $$


$$ S_n=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{ x}\right)^n $$

  • $\begingroup$ Thanks Jack :-) $\endgroup$ – tired Dec 15 '17 at 0:55
  • $\begingroup$ @tired: Thank you very much for the nice proof. It contains however a misprint. $4$ shall be replaced with $4^{1/n}$ in the denominator of the residue value. But probably simpler and more correctly were to write it just as $$\text{res}(f(z),z=z_0)=\frac{1}{4\sqrt{1-x^2}} \left(2\frac{1-\sqrt{1-x^2}}{x^2}\right)^n.$$ $\endgroup$ – user Dec 15 '17 at 11:56
  • $\begingroup$ @user355705 you are absolutly right $\endgroup$ – tired Dec 16 '17 at 15:07

Extracting coefficients on the RHS we get the integral (coefficient on $x^{n+2k}$)

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+2k+1}} \frac{1}{\sqrt{1-z^2}} \left(\frac{1-\sqrt{1-z^2}}{z}\right)^n \; dz.$$

Now we put $(1-\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has $w = \frac{1}{2} z + \cdots$ so the image in $w$ of the contour in $z$ can be deformed to a small circle enclosing the origin in the $w$-plane. (Moreover we see that the exponentiated term starts at $z^n$ which justifies the corresponding offset in the series.) We get $dz = 2/(1+w^2) - 4w^2/(1+w^2)^2 \; dw = 2(1-w^2)/(1+w^2)^2 \; dw.$ We also have $1-z^2 = 1 - 4w^2/(1+w^2)^2 = (1-w^2)^2/(1+w^2)^2.$ All of this yields

$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k+1}}{2^{n+2k+1} w^{n+2k+1}} \frac{1}{(1-w^2)/(1+w^2)} w^n \frac{2(1-w^2)}{(1+w^2)^2} \; dw \\ = \frac{1}{2^{n+2k}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k}}{w^{2k+1}} \; dw.$$

This evaluates by inspection to

$$\frac{1}{2^{n+2k}} [w^{2k}] (1+w^2)^{n+2k} = \frac{1}{2^{n+2k}} [w^{k}] (1+w)^{n+2k} \\ = \frac{1}{2^{n+2k}} {n+2k\choose k}$$

which is the claim.


I plan to insert this approach into the next (2019) version of my notes.
The even powers of $\arcsin(x)$ all have a nice Maclaurin series (see eqs (19),(20),(21),(22)), and this can be seen as a side effect of the Lagrange–Bürmann inversion theorem. I am going to implement the same trick here, by computing the derivatives at the origin of the RHS (the opposite way has already been investigated by tired).

$$[x^m]\left[\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n\right]=\frac{1}{2\pi i}\oint\frac{(1-\sqrt{1-x^2})^n}{x^{n+m+1}\sqrt{1-x^2}}\,dx$$ where the integral is performed along a small circle enclosing the origin. What about enforcing the substitution $x=\sin z$? The sine function is holomorphic and injective in a neighbourhood of the origin, hence the RHS is simply converted into $$ \frac{1}{2\pi i}\oint\frac{(1-\cos z)^n}{\left(\sin z\right)^{n+m+1}}\,dz =\frac{1}{2\pi i}\oint\frac{2^n \sin^{2n}\frac{z}{2}}{2^{n+m+1}\sin^{n+m+1}\frac{z}{2}\cos^{n+m+1}\frac{z}{2}}\,dz$$ or $$\frac{1}{2\pi i}\oint\frac{\tan^{n-m-1}z}{2^{m}\cos^{2m}z}\,dz=\frac{1}{2^{m+1}\pi i}\oint u^{n-m-1}(1+u^2)^{m-1}\,du$$ via $z\to 2z$ and $z\to\arctan u$. Now the RHS is trivially given by a binomial coefficient multiplied by $\frac{1}{2^m}$ (provided by $m\geq n$) and the claim is proved.


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.