Question

`\rm\sum_(n = 0)^ \infty \bigg( - \frac{1}2 \bigg)^(n) \frac{ \Gamma ( (1 + n)/(2) )}{ \Gamma(1 + \frac{n}2) } \\`​

Answer 1

Let
`S` be the sum. Introduce a factor of
`\Gamma\left(\frac12\right)=\sqrt\pi` to rewrite the summand as a beta function integral. Then in the sum, interchange it with the integral, and
`S` reduces to a simple integral.

`\displaystyle S = \sum_(n=0)^\infty \left(-\frac12\right)^n \frac{\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(\frac n2+1\right)}`

`\displaystyle . ~~ = \frac1{\sqrt\pi} \sum_(n=0)^\infty \left(-\frac12\right)^n \, \mathrm{B}\left(\frac n2+\frac12, \frac12\right) \\\\ ~~~~ = \frac1{\sqrt\pi} \sum_(n=0)^\infty \left(-\frac12\right)^n \int_0^1 (t^(n/2))/(\sqrt t √(1-t)) \, dt \\\\ ~~~~ = \frac1{\sqrt\pi} \int_0^1 (dt)/(\sqrt t √(1-t)) \sum_(n=0)^\infty \left(-\frac{\sqrt t}2\right)^n \\\\ ~~~~ = \frac2{\sqrt\pi} \int_0^1 (dt)/(\sqrt t√(1-t)\left(\sqrt t+2\right))`

Substitute
`u=\sqrt t+2` and
`du=(dt)/(2\sqrt t)`.

`\displaystyle S = \frac4{\sqrt\pi} \int_2^3 (du)/(u√(-(u-1)(u-3)))`

Now substitute
`u=1+\frac2{1+t^2}` and
`du=-(4t)/((1+t^2)^2)\,dt` - this comes from an Euler substitution of the form
`√(a(x-\alpha)(x-\beta))=(x-\alpha)t`.
`S` reduces drastically to a trivial arctangent integral.

`\displaystyle S = \frac8{\sqrt\pi} \int_0^1 (dt)/(t^2+3) \\\\ ~~~~ = \frac8{\sqrt\pi} \cdot \frac\pi{6\sqrt3} = \boxed{\frac43√(\frac\pi3)}`