Why doesn't $\sinh(x)$'s Taylor Series divide by half? Learning about Taylor Series, I have the problem sinh(x). Obviously, $\sinh(x) = \dfrac{e^x - e^{-x}}2$. I basically did it all correctly, since most of it cancels when compared to $e^x$'s taylor series. But that $\cfrac 12$ that's in the problem is throwing me off. Why isn't the summation series:
$$\sum 2\frac{x^{2x+1}}{(2n + 1)!} $$
That (* 2) on the bottom is what is confusing me. How is that getting cancelled out? All of the positive terms of n are getting cancelled out in e$^x$'s taylor series, but we still have these left. And they're still being divided by 2. 
What I'm getting for the series itself written out, is:
$\cfrac 12\left[x + \cfrac {x^3}{3!} + \cfrac{x^5}{5!}\right]$
Why is that 2 just forgotten about?
 A: Write out the Taylor series for $e^x$ and $e^{-x}$:
$$
\begin {align*}
e^x &= \sum_{n=0}^\infty \frac{x^n}{n!} \\
e^{-x} &= \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!}
\end {align*}
$$
Subtract them and you get
$$ e^x - e^{-x} = \sum_{n=0}^\infty \frac{1-(-1)^n}{n!} \, x^n $$
When $n$ is even, the coefficient is zero, and when $n$ is odd, it is $\frac{2}{n!}$. So the $\frac{1}{2}$ cancels this 2.
A: Write out the first few terms of each series
$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} \cdots $$
$$ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \cdots $$
Subtract the two series and you get
$$ e^x - e^{-x} = 2x + \frac{2x^3}{3!} + \frac{2x^5}{5!} + \cdots $$
Notice how each odd power coefficient doubles? Divide the above by $2$ and you obtain the $\sinh x$ series
A: The following manipulation can be proven valid as all of the series are absolutely convergent on all of $\mathbb{R}$ (actually on all of $\mathbb{C}$).
\begin{align}
\sinh x&=\frac{e^x - e^{-x}}{2} \\
&=\frac{1}{2}\left[\sum_{n\in\mathbb{N}}\frac{x^n}{n!} - \sum_{n\in\mathbb{N}} \frac{(-1)^nx^n}{n!}\right] \\
&=\frac{1}{2} \sum_{n\in\mathbb{N}} \left(1-(-1)^n\right) \frac{x^n}{n!}
\end{align}
Now, notice that when $n$ is even, then the summand will vanish. When $n$ is odd, there will be a factor of $2$. So,
\begin{align}
\sinh x &= \frac{1}{2}\sum_{n\in 2\mathbb{N}+1}2\cdot \frac{x^n}{n!} \\
&= \sum_{n\in 2\mathbb{N}+1} \frac{x^n}{n!} \\
&= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}
\end{align}
