What is the value of this series?

Question

$$\sum_{k=0}^\infty\frac{(-1)^k\left(e^{4k}+1\right)}{(2k+1)!}$$

I do know that $$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

I don't know how to continue though...

Answer 1

Your series is the sum of $2$ series:$$\sum_{k=0}^\infty\frac{(-1)^ke^{4k}}{(2k+1)!}=\sum_{k=0}^\infty\frac{(-1)^k(e^4)^k}{(2k+1)!}\tag1$$and$$\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}.\tag2$$Since $(1)=\sin(e^4)$ and $(2)=\sin(1)$, the answer is $\sin(e^4)+\sin(1)$.