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QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true? $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$

POSTSCRIPT. I like to record this presentable form by Alexander Burstein: $$\sum_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$

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  • $\begingroup$ This can be rewritten as $$\sum_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}}{2^{2n+2}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac{1}{x},$$ or equivalently, $$\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{2^{2n}\,(n+x)}=\frac{2^{2x}}{x\,\binom{2x}x}.$$ $\endgroup$ Jul 26 at 3:44
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    $\begingroup$ This is elementary textbook material. See exercise 8 to chapter 12 of Whittaker and Watson. $\endgroup$
    – Nemo
    Jul 26 at 10:28
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    $\begingroup$ The sum is $x^{-1}(\,{}_2F_1(1/2,𝑥;1+𝑥\mid 1)−1)$, which can be evaluated by Gauss's theorem, one of the standard hypergeometric summation theorems. Max Alekseyev's proof is essentially a special case of the usual proof of Gauss's theorem. $\endgroup$
    – Ira Gessel
    Jul 26 at 13:23
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    $\begingroup$ @Nemo: One of my tags is "soft question". Nonetheless, look at the rich and pedagogical responses. $\endgroup$ Jul 26 at 13:28
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    $\begingroup$ @T.Amdeberhan I wonder if there is a nice probabilistic interpretation of $$\sum_{n=0}^{\infty}{\frac{\binom{2n}{n}\binom{2x}{x}x}{2^{2(n+x)}(n+x)}}=1.$$ $\endgroup$ Jul 27 at 5:43
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\begin{align}\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}&= \int_0^1\sum_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}y^{n+x}}{2^{2n+2}}\,{\rm d}y\\&=\int_0^1 y^{x-1}\big((1-y)^{-1/2}-1\big){\rm d}y\\&=B\left(x,\frac12\right)-\frac1x\end{align} and the rest follows from the properties of beta function.

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  • $\begingroup$ Simple but clever. $\endgroup$ Jul 26 at 6:39
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Here is a method to calculate this yourself. I am omitting some of the details as it requires integrating certain integrals (which you can do by the change of variables).

We start from the power series $c(y)=\sum_nC_ny^n$ where $C_n$ is the $n$-the Catalan number. It is known this is equal to $\frac{1-\sqrt{1-4y}}{2y}$. This follows from the recursive relation of the Catalan numbers.

Now we have $$yc(y^2)=\sum_nC_ny^{2n+1}$$ We differentiate this to get the following: $$c(y^2)+2y^2c'(y^2)=\sum_n{2n+1\choose n+1}y^{2n}\implies c(y)+2yc'(y)=\sum_n{2n+1\choose n+1}y^{n}$$ Multiply this by $y^x$ and then integrate with respect to $y$: $$\int (c(y)y^x+2y^{x+1}c'(y))dy+Const=\sum_n({2n+1\choose n+1}/(n+x+1))y^{n+x+1}$$ Now you just need to calculate the integral (which is possible with some change of variables), after that just plug in $y=\frac{1}{2^2}$ and the divide the whole thing by $\frac{1}{2^{2x+1}}$ you should get the answer.

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Let, $f(x)=\frac{\sqrt{\pi}}{\Gamma(x)}$, The values of $f(x)$ at $x=-1,-2,...-N$ points are $y_{-n}=\frac{\binom{2n}{n}}{(-4)^n}$.

Now, for $N+1$ points $n=0,-1,-2,-3,..,-N$ we define $F_N(x):=\frac{\sqrt{\pi}}{\Gamma(x+\frac{1}{2})}N^x$ and $y_N(-n)=\frac{n!\binom{2n}{n}N^{-n}}{(-4)^n}$.

Hence, from Lagrange's interpolation:

$W(x)\sum_{n=0}^{N} \frac{1}{(n+x)(-1)^nn!(N-n)!}y_N(-n)≈F_N(x)$

[ Here, $W(x)=\prod_{n=0}^{N} (x+n)$ ]

or, $\frac{W(x)}{N!}\sum_{n=0}^{N} \frac{\binom{2n-1}{n}}{2^{2n-1}}N(N-1)..(N-n+1)N^{-n}≈F_N(x)$

[$2\binom{2n-1}{n}:=1$ as we get from the previous step]

Now, tending $N \to \infty$, and using $\lim\limits_{N \to \infty} \frac{W(x)}{N!N^x}=\frac{1}{\Gamma(x)}$

Hence, we get $\frac{1}{\Gamma(x)}\left(\frac{1}{x}+\sum_{n=0}^{\infty} \frac{\binom{2n+1}{n+1}}{2^{2n-1}(n+1+x)}\right)=\frac{\sqrt{\pi}}{\Gamma(x+\frac{1}{2})}$

The error term would go to zero as $N$ is increased when $x>-N$. This proves the identity.

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Q: Is this identity true?

A: Yes, Mathematica evaluates it as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP.

The identity holds for all real $x$ unequal to a negative integer. It holds in particular for $x=0$, when the sum equals $2\ln 2$.

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    $\begingroup$ Both sides are meromorphic with the same simple poles and residues, so their difference is entire after removing singularities. Standard calculations show that this difference is bounded on circles of half-integer radius (one can also use squares or rectangles if one prefers), so from Cauchy's inequalities the difference is constant (cf. Liouville's theorem). Now take the limit as $x \to +\infty$ to see that the difference in fact vanishes. $\endgroup$
    – Terry Tao
    Jul 25 at 17:49

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