Sum of the series $\sum_{n \ge 0}{\frac{x^{4n+1}}{(4n+1)!}}$

Question

I want to determine the sum of the series $$\sum_{n \ge 0}{\frac{x^{4n+1}}{(4n+1)!}}$$ I know this has to do with the sum $$\sum_{n \ge 0}{\frac{x^{n}}{(n)!}}=e^x\;\; \forall x\in \mathbb R$$ But i can't see how to start. Thank you for your help!!

Answer 1

Recall that $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ and $$\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots.$$ Add, divide by $2$.

If the $\sinh$ function is unfamiliar, replace it by the equivalent $\frac{e^x-e^{-x}}{2}$. Use the power series for $e^t$ to find the power series for $\frac{e^{x}-e^{-x}}{2}$.