Evaluate the sum $\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$ $$\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$
I am having difficulty finding the function that represents this series. 
I have only found radius of convergence which is $(-\infty,+\infty)$ from the ratio test.
 A: Define $$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$$ hence $$f'(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$ and then 
$$f''(x)=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n-1)!}=f(x)$$
Now note that $f(0)=0$ and $f'(0)=1$. Therefore we need to solve $f''(x)-f(x)=0$. Using Laplace transforms we get $$s^2F(s)-sf(0)-f'(0)-F(s)=0$$ or that $$F(s)=\frac{1}{s^2-1}=\frac12\frac{1}{s-1}-\frac12\frac{1}{s+1}$$ and taking an inverse we obtain $$f(x)=\frac{e^x}{2}-\frac{e^{-x}}{2}=\sinh(x)$$
A: Let $f(x)$ be the function represented by your series. Then 
$$f(x) = \frac{1}{2}\sum_{n = 0}^\infty \frac{1 - (-1)^n}{n!}x^n$$
since $1 - (-1)^n$ is $0$ when $n$ is even and $2$ when $n$ is odd. Thus
$$f(x) =  \frac{1}{2}\sum_{n = 0}^\infty \frac{x^n}{n!} - \frac{1}{2}\sum_{n = 0}^\infty \frac{(-x)^n}{n!}, $$
which simplifies to 
$$f(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x}.$$
A: That is the Taylor series of $\sinh$, hyperbolic sine.
The derivation of this comes from:
$$D_x \sinh x = \cosh x$$
$$D_x \cosh x = \sinh x$$
and:
$$\sinh 0 = 0$$
$$\cosh 0 = 1$$
A: This happens to be precisely the odd terms of:
$$e^x=\sum_{n\ge0}\frac{x^n}{n!}$$

Let's be a bit more general. Let's say we have a function $A$, defined by:
$$A(x)=\sum_{n\ge0}a_nx^n$$
where $a_n$ is some sequence. How do we get only the odd terms? Let's call the function generated by the odd terms $A_O$. ("O" for "odd.") So, we want this function:
$$A_O(x)=\sum_{n\ge0}a_{2n+1}x^{2n+1}$$
Well, here's a little trick. Let's write out $A(x)$, like this:
$A(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\dotsb$
And, now, write out $A(-x)$:
$A(-x)=a_0-a_1x+a_2x^2-a_3x^3+a_4x^4-a_5x^5+\dotsb$
And, finally, find $A(x)-A(-x)$.
$A(x)-A(-x)=2a_1x+2a_3x^3+2a_5x^5+\dotsb$
because everything else cancels. This is just twice what we want! So:
$$A_O(x)=\frac{A(x)-A(-x)}2$$

And, to solve your own problem, we have:
$$\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}=\frac{e^x-e^{-x}}2$$
