An exercise from Grimmett & Stirzaker 3.1 question 2: For a random variable $X$ with a mass function on positive integers $$f(x)=\frac{1}{e^2-1}\frac{2^x}{x!},$$ what is the probability that $X$ is even?
We want $\mathbb{P}(\text{$X$ even})=\sum_{k=1}^{\infty}\mathbb{P}(X=2k)$. In the answer there is one equal sign I can't wrap my head around: $$\sum_{k=1}^{\infty}\frac{2^{2k}}{(2k)!}=\sum_{i=0}^{\infty}\frac{2^{i}+(-2)^{i}}{2(i)!}-1$$
If anyone can explain this to me, help is much appreciated
Note that: $$\sum_{k=1}^{\infty}\frac{2^{2k}}{(2k)!}=\sum_{i=1}^{\infty}\frac{2^i}{i!}χ(i)$$ where $χ(2i)=1$ and $χ(2i+1)=0$, for all $i\geq1$
Now we can write $χ$ in a closed form as follows:
$$χ(i)=\frac12 \left(1+(-1)^i\right)$$
Substituting this in the original expression, we get:
$$\sum_{i=1}^{\infty}\frac{2^i}{i!}χ(i)=\sum_{i=1}^{\infty}\frac{2^i\left(1+(-1)^i\right)}{2\cdot i!}=\sum_{i=1}^{\infty}\frac{2^i+(-2)^i}{2\cdot i!}$$
To get the final wanted equality, notice that for $i=0$ the summand gives $1$, and so we can add and subtract $1$ in order to make the sum start from $i=0$, giving us finally:
$$\sum_{k=1}^{\infty}\frac{2^{2k}}{(2k)!}=\sum_{i=0}^{\infty}\frac{2^i+(-2)^i}{2\cdot i!}-1$$