Prove the infinite sum $\sum_{k=0}^{\infty}{\frac{(2k-1)!!}{(2k)!!}x^{2k}}=\frac{1}{\sqrt{1-x^2}}$ I have been trying to prove that $$\sum_{k=0}^{\infty}{\frac{(2k-1)!!}{(2k)!!}x^{2k}}=\frac{1}{\sqrt{1-x^2}}$$
I have done it by using the binomial formula, bit we can't use the gamma function and putting $-\frac{1}{2}$ is kinda logical but not that clear. I have also tried to calculate the partial sum formula for the series but it gets complicated so I can't get something nice of it... is the some kind of other way to calculate it?
 A: The binomial method should work but here is an other method. Let's note $f(x) = \frac{1}{\sqrt{1-x^2}}$. Then by direct computation you can see that $$(1-x^2)f'(x) -xf(x)=0.$$ But if this equation has an analytic solution $\sum_{n=0}^{+\infty}a_n x^n$, then plugging the sum in the equation and solving the recursion problem on the $a_n$ gives $$\sum_{k=0}^{\infty}{\frac{(2k-1)!!}{(2k)!!}x^{2k}}$$ as solution.
A: Since $$\left(2k\right)!!=2^{k}k!
 $$ and $$\left(\frac{1}{2}\right)_{k}=\frac{\left(2k-1\right)!!}{2^{k}}
 $$ where $\left(x\right)_{n}=x\left(x+1\right)\cdots\left(x+n-1\right)
 $ is the Pochhammer symbol, we have $$\sum_{k\geq0}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}x^{2k}=\sum_{k\geq0}\frac{\left(\frac{1}{2}\right)_{k}}{k!}x^{2k}=\frac{1}{\sqrt{1-x^{2}}}
 $$ from the generalized binomial theorem.
A: $$\frac{(2k-1)!!}{((2k)!!}=\frac{1\cdot3\cdot5\cdot\ldots\cdot(2k-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2k)}=\frac{(2k)!}{(2\cdot4\cdot6\cdot\ldots\cdot(2k))^2}=$$
$$=\frac{(2k)!}{2^{2k}(k!)^2}\implies\text{we have the series}\;\;\sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2}x^{2k}$$
If you still cannot recognize it, as it is easy to see it converges for $\;|x|<1\;$ , integrate the series (elementwise within its convergence radius):$${}$$
$$\int\sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2}x^{2k}\;dx=\sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2}\int x^{2k}\,dx=$$$${}$$
$$=\sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2}\frac{x^{2k+1}}{2k+1}=\arcsin x$$
as you can check, for example here . Now just go back, differentiating, and you get what you want.
A: This problem is a nice example of one that can be solved in a variety of ways, and a nice illustration of the fact that the most natural approach can depend a great deal on one’s background. Just for fun, here’s yet another way, the one that occurred to me first. 
First note that
$$\frac{(2k-1)!!}{(2k)!!}=\frac{(2k)!}{(2k)!!^2}=\frac{(2k)!}{2^{2k}k!^2}=\binom{2k}k\left(\frac12\right)^{2k}\;,$$
so
$$\sum_{k\ge 0}\frac{(2k-1)!!}{(2k)!!}x^{2k}=\sum_{k\ge 0}\binom{2k}k\left(\frac{x^2}4\right)^k=\sum_{k\ge 0}(k+1)C_k\left(\frac{x^2}4\right)^k\;,$$
where $C_k=\frac1{k+1}\binom{2k}k$ is the $k$-th Catalan number. The generating function for the Catalan numbers is
$$c(x)=\sum_{k\ge 0}C_kx^k=\frac{1-\sqrt{1-4x}}{2x}\;;$$
a derivation can be found here, among many other places. Thus,
$$\frac12\left(1-\sqrt{1-4x}\right)=xc(x)=\sum_{k\ge 0}C_kx^{k+1}\;,$$
and we can differentiate to get
$$\frac1{\sqrt{1-4x}}=\sum_{k\ge 0}(k+1)C_kx^k\;.$$
Replacing $x$ by $\frac{x^2}4$, we find that
$$\sum_{k\ge 0}\frac{(2k-1)!!}{(2k)!!}x^{2k}=\sum_{k\ge 0}(k+1)C_k\left(\frac{x^2}4\right)^k=\frac1{\sqrt{1-x^2}}\;.$$
