Sum of infinite fraction

[man111]

The sum of the series [tex]\displaystyle \frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6 \cdot 8}+...............\bf{\infty}[/tex] terms

[nevrodermit]

We want to calculate [tex]s=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+\cdots=\frac{1}{2.2!}+\frac{1.3}{2^2.3!}+\frac{1.3.5}{2^3.4!}+\frac{1.3.5.7}{2^4.5!}+\cdots[/tex]
[tex](1-x)^{-\frac{1}{2}}=1+\frac{x}{2}+\frac{1.3}{2^2.2!}x^2+\frac{1.3.5}{2^33!}x^3+\frac{1.3.5.7}{2^44!}x^4 +\cdots[/tex]
[tex]\int{(1-x)^{-\frac{1}{2}}}dx=x+\frac{x^2}{4}+\frac{1.3}{2^2.3!}x^3+\frac{1.3.5}{2^34!}x^4+\frac{1.3.5.7}{2^45!}x^5+\cdots[/tex]
[tex]\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=1+\frac{1}{4}+\frac{1.3}{2^2.3!}+\frac{1.3.5}{2^3.4!}+\frac{1.3.5.7}{2^45!}+\cdots[/tex]
Therefore, [tex]s=\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx-1[/tex]

The integral is
[tex]\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=-\int_{0}^{1}(1-x)^{-\frac{1}{2}}d(1-x)=2(-\sqrt{1-x}|_1 +\sqrt{1-x}|_0)=2[/tex]
So [tex]s=1[/tex]

[man111]

Thanks nevrodermit for nice solution.