$\sum \limits_{k~ \in~ K_e} \frac{a^k}{k!} - \sum \limits_{k~ \in~ K_o} \frac{a^k}{k!} = \sum \limits_{k=0}^{\infty} \frac{(-a)^k}{k!}$

Question

Given:

$$K_e = \{\text{even numbers: 0, 2, 4, ...}\}$$

$$K_o = \{\text{odd numbers: 1, 3, 5, ...}\}$$

How to prove this equality is true?

$$\sum \limits_{k~ \in~ K_e} \frac{a^k}{k!} - \sum \limits_{k~ \in~ K_o} \frac{a^k}{k!} = \sum \limits_{k=0}^{\infty} \frac{(-a)^k}{k!}$$

There is nothing to prove... Since $\sum_{k=0}^\infty \frac{(-a)^k}{k!}$ is absolutely convergent you can permute the order of summing without changing the limit, and thus $$\sum_{k=0}^\infty \frac{(-a)^k}{k!}=\sum_{k=0}^\infty \frac{(-a)^{2k}}{(2k!)}+\sum_{k=0}^\infty \frac{(-a)^{2k+1}}{(2k+1)!}=...$$ — Surb, Mar 10 '20 at 14:17

score 1 · Answer 1 · answered Mar 10 '20 at 14:17

1

$(-a)^k = a^k$ for $k$ even and

$(-a)^k = - a^k$ for $k$ odd

Split the series in two and apply it

answered Mar 10 '20 at 14:17

Pablo Gregori

97

score 1 · Answer 2 · answered Mar 10 '20 at 14:18

$$\begin{aligned}\sum_{k\in\mathbb{K_e}}\frac{a^k}{k!}&=1+\frac{a^2}{2!}+\frac{a^4}{4!}+\cdots \\ \sum_{k\in\mathbb{K_o}}\frac{a^k}{k!}&=\frac{a}{1!}+\frac{a^3}{3!}+\frac{a^5}{5!}+\cdots\\ \sum_{k=0}^{\infty}\frac{(-1)^ka^k}{k!}&=1-\frac{a}{1!}+\frac{a^2}{2!}-\frac{a^3}{3!}+\cdots\end{aligned}$$

score 1 · Answer 3 · answered Mar 10 '20 at 14:18

It is true since $$ \sum\limits_{k \in K_e } {\frac{{a^k }}{{k!}}} = \sum\limits_{k \in K_e } {\frac{{( - a)^k }}{{k!}}} $$ and $$ - \sum\limits_{k \in K_o } {\frac{{a^k }}{{k!}}} = \sum\limits_{k \in K_o } {( - 1)\frac{{a^k }}{{k!}}} = \sum\limits_{k \in K_o } {( - 1)^k \frac{{a^k }}{{k!}}} = \sum\limits_{k \in K_o } {\frac{{( - a)^k }}{{k!}}} . $$

$\sum \limits_{k~ \in~ K_e} \frac{a^k}{k!} - \sum \limits_{k~ \in~ K_o} \frac{a^k}{k!} = \sum \limits_{k=0}^{\infty} \frac{(-a)^k}{k!}$

3 Answers3