\begin{align*} \lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{B\left(n-k+1,k+1\right)\left(2k+1\right)} & =\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}\left(n-k+1\right)\left(k+1\right)C\left(n+2,n-k+1\right)}{\left(n+2\right)\left(2k+1\right)}\cmt{B\left(x,y\right)=\frac{x+y}{xyC\left(x+y,x\right)}}\\ & =\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}\left(n-k+1\right)\left(k+1\right)C\left(n+2,k+1\right)}{\left(n+2\right)\left(2k+1\right)}\\ & =\lim_{n\rightarrow\infty}\frac{n+1}{\sqrt{n}}\sum_{k=0}^{n}C\left(n,k\right)\frac{\left(-1\right)^{k}}{\left(2k+1\right)}\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\sum_{k=0}^{n}C\left(n,k\right)\left(-1\right)^{k}\left[\frac{1}{2k+1}x^{2k+1}\right]_{-1}^{1}\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^{1}\sum_{k=0}^{n}C\left(n,k\right)\left(-1\right)^{k}x^{2k}dx\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^{1}\sum_{k=0}^{n}C\left(n,k\right)\left(-x^{2}\right)^{k}dx\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^{1}\left(1-x^{2}\right)^{n}dx\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\int_{-1}^{1}\left(1+x\right)^{n}\left(1-x\right)^{n}dx\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}B\left(n+1,n+1\right)2^{n+n+1}\cmt{\because\int_{-1}^{1}\left(1+t\right)^{x}\left(1-t\right)^{y}dt=2^{x+y+1}B\left(x+1,y+1\right)}\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma^{2}\left(n+1\right)}{\Gamma\left(2\left(n+1\right)\right)}2^{2n+1}\cmt{\because B\left(x,y\right)=\frac{\Gamma\left(x\right)\Gamma\left(y\right)}{\Gamma\left(x+y\right)}}\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\sqrt{\pi}\Gamma^{2}\left(n+1\right)}{2^{2\left(n+1\right)-1}\Gamma\left(n+1\right)\Gamma\left(n+1+\frac{1}{2}\right)}2^{2n+1}\cmt{\because\Gamma\left(2z\right)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{2}\right)}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+1+\frac{1}{2}\right)}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+1+\frac{1}{2}\right)}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+1+\frac{1}{2}\right)}\sqrt{n+\frac{1}{2}}\frac{\Gamma\left(n+1+\frac{1}{2}\right)}{\Gamma\left(n+2\right)}}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\sqrt{\frac{\sqrt{n}\sqrt{n+\frac{1}{2}}}{\left(n+1\right)}}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\left(\frac{n^{2}+\frac{1}{2}n}{\left(n+1\right)^{2}}\right)^{\frac{1}{4}}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\left(\frac{\left(n+1\right)^{2}-\frac{3}{2}n-1}{\left(n+1\right)^{2}}\right)^{\frac{1}{4}}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\left(1-\frac{\frac{3}{2}n+1}{\left(n+1\right)^{2}}\right)^{\frac{1}{4}}\\ & =\frac{\sqrt{\pi}}{2} \end{align*}