Summation relating factorial and cosine

Question

How to simplify \begin{align*} \sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(2k\right)!}{4^{k}\left(k!\right)^{2}}\cos\left(kx\right) \end{align*}

for $0\leq x <\pi$ ? I don't even know where to start.

Answer 1

If you know that $$ \sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(2k\right)!}{4^{k}\left(k!\right)^{2}}x^k=\frac{1}{\sqrt{1+x}} $$ the problem can be solved in the following manner.

Consider $$I= \sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(2k\right)!}{4^{k}\left(k!\right)^{2}}\cos\left(kx\right)$$ $$J= \sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(2k\right)!}{4^{k}\left(k!\right)^{2}}\sin\left(kx\right)$$ $$I+iJ=\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(2k\right)!}{4^{k}\left(k!\right)^{2}}e^{ikx}=\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(2k\right)!}{4^{k}\left(k!\right)^{2}}\left(e^{ix}\right)^k=\frac{1}{\sqrt{1+e^{i x}}}$$

I am sure that you can take it from here.