Prove that $\sum_{n=0}^\infty {\frac{x^{2n+1}}{(2n+1)!}} = \frac{e^x - e^{-x}}{2}$

Question

I have been trying to show:

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

I have come so far as to show:

$\begin{aligned} \sum_{n=0}^{\infty} {\frac{x^{2n+1}}{(2n+1)!}} &= \frac{x^1}{1!} + \frac{x^3}{3!} + \cdots \\ &= \left(\frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \right) - \left(\frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \right) \\ &= e^x - \left(\frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \right) \end{aligned}$

Any advice on how to proceed would be much appreciated!

Answer 1

\begin{align}\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}&=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\\&=\frac12\left(2x+2\frac{x^3}{3!}+2\frac{x^5}{5!}+\cdots\right)\\&=\frac12\left(\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)-\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots\right)\right)\\&=\frac{e^x-e^{-x}}2.\end{align}

Answer 2

Of course, as commented, begin with the right side:

$$\frac12(e^x-e^{-x})=\frac12\sum_{n=0}^\infty\frac{[1-(-1)^n]x^n}{n!}=(**)$$

But since

$$1-(-1)^n=\begin{cases}0,&n\;\text{is even}\\{}\\2,&n\;\text{is odd}\end{cases}$$

We get:

$$(**)=\frac12\sum_{n=0}^\infty\frac{2x^{2n+1}}{(2n+1)!}$$

Answer 3

The constant term is missing in the expansion of $\mathrm e^x$. This being said, if you want to start from the l.h.s., here is a way: \begin{align} 2\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}&=\sum_{n=0}^\infty \biggl(\frac{x^{2n}}{(2n)!}+\frac{x^{2n+1}}{(2n+1)!}\biggr)-\sum_{n=0}^\infty\biggl(\frac{x^{2n}}{(2n)!}-\frac{x^{2n+1}}{(2n+1)!}\biggr) \\ &=\sum_{m=0}^\infty \frac{x^{m}}{m!}-\sum_{n=0}^\infty\biggl(\frac{(-x)^{2n}}{(2n)!}+\frac{(-x)^{2n+1}}{(2n+1)!}\biggr)\\[1ex] &=\sum_{m=0}^\infty \frac{x^{m}}{m!}-\sum_{m=0}^\infty \frac{(-x)^{m}}{m!}=\mathrm e^x-\mathrm e^{-x}. \end{align}