If $f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!!}$ then prove that $f'(x)=1+xf(x)$ 
Question If $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!!}$
  then prove that
$f'(x)=1+xf(x)$ 

Book's Question

I can prove its radius of convergence is infinity
Book's Answer

My Approach
$\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!!}$ 
$\displaystyle f'(x)=\sum_{n=1}^{\infty}\frac{(2n+1)x^{2n}}{(2n+1)!!}$ 
Now I cannot prove $f'(x)=1+xf(x)$ 
 A: Hints:


*

*In your derivation, what happens to the $n=0$ term?  It is not a constant.  In particular, for $n=0$, the corresponding term in $f$ is $\frac{x}{1!!}=x$, so its derivative is $1$.

*Consider $\frac{2n+1}{(2n+1)!!}$.  Since the double factorial is
$$(2n+1)!!=(2n+1)(2n-1)(2n-3)\dots1,$$
the first factor of the double factorial cancels, leaving $(2n-1)!!$.  Note that $!!$ does not mean taking the factorial twice.
A: If you are given a power series
$$
f(x)=\sum_{n=0}^{\infty}a_nx^n
$$
with positive (possibly infinite) radius of convergence, then in the interval of convergence
$$
f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}\tag{*}
$$
The shift in the starting index is done because $x^{n-1}$ would be wrong for $n=0$.
However, your series is not in the form (*), because it has no term for $x^0$, so we don't need to take care of it when differentiating term by term. Hence
$$
f'(x)=\sum_{n=0}^{\infty}\frac{(2n+1)x^{2n}}{(2n+1)!!}
$$
Now we can only simplify
$$
\frac{2n+1}{(2n+1)!!}=\frac{1}{(2n-1)!!}
$$
when $n>0$, so we have to detach the initial term:
$$
f'(x)=1+\sum_{n=1}^{\infty}\frac{(2n+1)x^{2n}}{(2n+1)!!}
=1+\sum_{n=1}^{\infty}\frac{x^{2n}}{(2n-1)!!}
=1+x\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n-1)!!}
$$
with the shift $n=k+1$ we finally get
$$
f'(x)=1+x\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)!!}
=1+xf(x)
$$
Caveat
The definition of the semifactorial $m!!$ is
$$
m!!=
\begin{cases}
2\cdot 4\cdot \ldots \cdot m & \text{$m$ even} \\[4px]
1\cdot 3\cdot \ldots \cdot m & \text{$m$ odd}
\end{cases}
$$
