Sum of Series

[man111]

Sum of [tex]n[/tex] terms of the series [tex]\displaystyle \frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+..[/tex]

[drago]

@man111: Could you be more careful with the wording and the meaning of the problems you post. In this case it should be:

Find the sum of the series [tex]\displaystyle S=\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\dots[/tex]

Making some calculations we get that:
\[S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(2n)!}{n! (n+1)!}4^{-n} \]
Now, note that $C_n=\frac{(2n)!}{n! (n+1)!}$ is exactly the $n$-th Catalan number. It's known the generating function of the Catalan numbers is $C(x)=\frac{1-\sqrt{1-4x}}{2x}$ (see here).That is:
\[C(x)=\sum_{n=0}^{\infty}\frac{(2n)!}{n! (n+1)!}x^n\]
The series is converging for $|x|\leq 1/4$, so pluging in $x=1/4$, we get:
\[S=1\]

[man111]

Thanks Drago.