Proving a binomial sum identity

Question

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: $$\sum_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$

Answer 1

\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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $2\ln 2$.