What is the finite sum of an alternating odd sequence with binomial coefficients: $ \ \sum_{k=0}^{j} \binom{j}{k} \ \frac{(-1)^k}{2k+1}$?

Question

Do we have a nice expression for the following finite series? $$ \sum_{k=0}^{j} \binom{j}{k} \ \frac{(-1)^k}{2k+1}.$$ Any help is really appreciated since I can skip the part concerning the partial sums.

Answer 1

We have $$ \frac{1}{2k+1} = \int_0^1 \frac{1}{2} x^{k-1/2} \, dx, $$ so interchanging this integral with the sum, $$ \sum_{k=0}^j \binom{j}{k} (-1)^k \int_0^1 \frac{1}{2} x^{k-1/2} \, dx = \frac{1}{2}\int_0^1 x^{-1/2} \sum_{k=0}^j \binom{j}{k} (-x)^k \, dx = \frac{1}{2}\int_0^1 x^{-1/2} (1-x)^j \, dx. $$ This is a special case of the Beta-function, or we can write it as $$ \frac{1}{2}\int_0^1 x^{-1/2} (1-x)^j \, dx = \binom{j+1/2}{j}^{-1}, $$ which of course expands to $$ \frac{j!}{(j+1/2)(j-1/2) \dotsm (5/2) (3/2)} = \frac{2^j j!}{(2j+1)!!}. $$

Answer 2

An overkill. From the Melzak's identity $$\sum_{k=0}^{j}\dbinom{j}{k}\left(-1\right)^{k}\frac{f\left(y-k\right)}{k+x}=\frac{f\left(x+y\right)}{x\dbinom{j+x}{j}}$$ where $f$ is an algebraic polynomial up to degree $j$ and $x\neq-k$ we get, taking $f\left(y\right)\equiv1$ and $x=1/2$, that $$\frac{1}{2}\sum_{k=0}^{j}\dbinom{j}{k}\frac{\left(-1\right)^{k}}{k+1/2}=\color{red}{\dbinom{j+1/2}{j}^{-1}}$$