Show that $ \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2)$.

Question

I want to prove that $$ \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2),$$

where $B(\cdot , \cdot)$ is the beta function.

My idea was to change it to something like my previous question.

Edit 1.It follows from the absorption formula that $$ \binom{2j+1}{k+j+1} = \frac{(2j+1)(2j) \ldots(j+2)(j+1)}{(k+j+1)(k+j) \ldots(k+2)(k+1)} \binom{j}{k}.$$ How can I go further with this binomial series?

Edit 2. This sum is going to diverge very fast as $j \to \infty,$ so I guess something like $\binom{2j}{j}$ inolves.

Edit 3. Due to $(-1)^k$, we have got lots of cancellations, which makes the series to be controlled.

Edit 4. The problem still open.

Edit 5. [Getting some progress] Consider two polynomials $$p_j(t):= \sum_{k=0}^{j} \binom{2j+1}{k+j+1} (-t)^k \ \, \text{and} \ \ q_j(t):=4^j \ (1-t)^{j}.$$

To prove our guess, it suffices to show that $$ \color{red}{\int_0^1 t^{-1/2} \, p_j(t) \, dt = \int_0^1 t^{-1/2} \, q_j(t) \, dt} \tag{*}$$ since $$ \begin{align} \sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1} & = \sum_{k=0}^{j} \binom{2j+1}{k+j+1} (-1)^k \int_0^1 \frac{1}{2} t^{k-1/2} \, dt \\ & = \frac{1}{2} \int_0^1 t^{-1/2} \, p_j(t) \, dt \, , \end{align} $$ and $$ \frac{1}{2} \int_0^1 t^{-1/2} \, q_j(t) \, dt= 2^{2j-1}\ B(j+1,\frac{1}{2}).$$ I guess we can use induction since for $j=1$, we have $$ \int_0^1 t^{-1/2} \, (3-t) \, dx = \int_0^1 t^{-1/2} \, 4(1-t) \, dt = 16/3.$$

Answer 1

Numerically it is seems that

$$\sum_{k=0}^{j} \binom{2j+1}{k+j+1} \ \frac{(-1)^k}{2k+1}=2^{2j-1}\ B(j+1,1/2),$$

where $B(\cdot , \cdot)$ is the beta function. I will update the progress in the original post.

Answer 2

$S_n=\sum\limits_{k=0}^{n} \dbinom{2n+1}{k+n+1} \ \dfrac{(-1)^k}{2k+1}$

where $j=n$ (I am sorry I used $n$ instead of $j$ during the long proving).

1,

$\dbinom{2n+1}{k+n+1}=\dfrac{(2n+1)!}{(n+k+1)!(n-k)!}\dfrac{n!}{k!}\dfrac{k!}{n!}\dfrac{(n+1)!}{(n+1)!}=$ $\dbinom{n}{k}\dfrac{\dbinom{2n+1}{n}}{\dbinom{n+k+1}{k}}$

So we have:

$S_n=\sum\limits_{k=0}^{n} \dbinom{n}{k}\dfrac{\dbinom{2n+1}{n}}{\dbinom{n+k+1}{k}} \dfrac{(-1)^k}{2k+1}=\dbinom{2n+1}{n}(n+1)\sum\limits_{k=0}^{n} \dbinom{n}{k}\dfrac{\Gamma(k+1)\Gamma(n+1)}{\Gamma(n+k+2)}\dfrac{(-1)^k}{2k+1}$

2, Using that $\beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

$S_n=\dbinom{2n+1}{n}(n+1)\sum\limits_{k=0}^{n} \dbinom{n}{k}\beta(n+1,k+1)\dfrac{(-1)^k}{2k+1}$

3,

$(-1)^k\dfrac{1}{2k+1}=\int\limits_0^1 (-t^2)^kdt$

and

$\beta(k+1,n+1)=\int\limits_0^1 u^k (1-u)^n du$ by the definition of beta function. Put them back into the sum, and rearrange:

$S_n=\binom{2n+1}{n}(n+1)\sum\limits_{k=0}^{n} \binom{n}{k}\int\limits_0^1u^k (1-u)^n du\int\limits_0^1 (-t^2)^kdt=$

$\binom{2n+1}{n}(n+1)\int\limits_0^1 (1-u)^n \int\limits_0^1\sum\limits_{k=0}^{n}\binom{n}{k}(-ut^2)^kdtdu$

4,

As $\sum\limits_{k=0}^{n}\binom{n}{k}(-ut^2)^k=(1-ut^2)^n$

$S_n=\binom{2n+1}{n}(n+1)\int\limits_0^1(1-u)^n\int\limits_0^1 (1-ut^2)^ndtdu$

Using the $ut^2=x$ substitution we get:

$S_n=\frac{1}{2}\binom{2n+1}{n}(n+1)\int\limits_0^1(1-u)^n u^{-\frac{1}{2}}\int\limits_0^u (1-x)^n x^{-\frac{1}{2}}dtdu$

5,

Applying the definitions of beta and incomplete beta function:

$S_n=\frac{1}{2}\binom{2n+1}{n}(n+1)\int\limits_0^1(1-u)^n u^{-\frac{1}{2}}\beta(u;n+1,\frac{1}{2})du$

There is the function and its derivate in the integral $(\int f'(x)f(x)dx=\frac{f^2(x)}{2})$ so

$S_n=\frac{1}{4}\binom{2n+1}{n}(n+1)\beta^2(n+1,\frac{1}{2})$

6,

Easy to prove that $\frac{1}{2}\binom{2n+1}{n}(n+1)\beta(n+1,\frac{1}{2})=2^{2n}$

Finally

$S_n=2^{2n-1}\beta(n+1,\frac{1}{2})$