Evaluate $\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!}$ I want to evaluate $\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!}$ but while doing my research, I noticed that
\begin{align*}
&\sin(z) = \sum^{\infty}_{n=1} (-1)^n \frac{z^{2n+1}}{(2n+1)!} \\
&\sinh(z) = \sum^{\infty}_{n=1} \frac{z^{2n+1}}{(2n+1)!}
\end{align*}
So
\begin{align*}
\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!} = \frac{\sin(z)+\sinh(z)}{2}
\end{align*}
Is it that easy?
 A: Consider the following.
\begin{align}
\frac{1}{\sqrt{2}} \, \cosh\left(\frac{x}{\sqrt{2}}\right) \, \sin\left(\frac{x}{\sqrt{2}}\right) &= \sum_{r} \frac{x^{2 r}}{2^{r} \, (2 r)!} \times \sum_{s} \frac{(-1)^s \, x^{2 s + 1}}{2^{s+1} \, (2 s + 1)!} \\
&= \sum_{n} \sum_{s=0}^{n} \frac{(-1)^s \, x^{2 n +1}}{2^{n+1} \, (2n - 2s)! \, (2 s + 1)!} \\
&= \sum_{n} \left( \sum_{s=0}^{n} (-1)^s \, \binom{2 n +1}{2 s+1} \right) \, \frac{x^{2 n +1}}{2^{n+1} \, (2 n+1)!} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} (-1)^{\lfloor{n/2\rfloor}} \, \frac{x^{2 n+1}}{(2 n+1)!}.
\end{align}
In a similar manor it can be found that
$$ \frac{1}{\sqrt{2}} \, \cos\left(\frac{x}{\sqrt{2}}\right) \, \sinh\left(\frac{x}{\sqrt{2}}\right) = \frac{1}{2} \, \sum_{n=0}^{\infty} (-1)^{\lfloor{n/2\rfloor}} \, \frac{(-1)^n \, x^{2 n+1}}{(2 n+1)!}. $$
Now
\begin{align}
S &= \frac{1}{\sqrt{2}} \, \cosh\left(\frac{x}{\sqrt{2}}\right) \, \sin\left(\frac{x}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \, \cos\left(\frac{x}{\sqrt{2}}\right) \, \sinh\left(\frac{x}{\sqrt{2}}\right) \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} (-1)^{\lfloor{n/2\rfloor}} \, \frac{(1 + (-1)^n) \, x^{2 n+1}}{(2 n+1)!} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n \, x^{4 n+1}}{(4 n+1)!}.
\end{align}
A different series can be obtained from:
\begin{align}
\frac{1}{2} \left( \sin(x) + \sinh(x)\right) &= \sum_{n=0}^{\infty} \frac{1 + (-1)^n}{2} \, \frac{x^{2n+1}}{(2n+1)!} \\
&= \sum_{n=0}^{\infty} \frac{x^{4 n+1}}{(4 n +1)!}
\end{align}
