How can we show that $\sum_{n=0}^{\infty}{2n^2-n+1\over 4n^2-1}\cdot{1\over n!}=0?$ Consider

$$\sum_{n=0}^{\infty}{2n^2-n+1\over 4n^2-1}\cdot{1\over n!}=S\tag1$$
  How does one show that $S=\color{red}0?$

An attempt:
$${2n^2-n+1\over 4n^2-1}={1\over 2}+{3-2n\over 2(4n^2-1)}={1\over 2}+{1\over 2(2n-1)}-{1\over (2n+1)}$$
$$\sum_{n=0}^{\infty}\left({1\over 2}+{1\over 2(2n-1)}-{1\over (2n+1)}\right)\cdot{1\over n!}\tag2$$
$$\sum_{n=0}^{\infty}\left({1\over 2n-1}-{2\over 2n+1}\right)\cdot{1\over n!}=\color{blue}{-e}\tag3$$
Not sure what is the next step...
 A: Hint:
\begin{eqnarray}
&&\sum_{n=0}^{\infty}{2n^2-n+1\over 4n^2-1}\cdot{1\over n!}\\
&=&\sum_{n=0}^{\infty}{(2n^2+n)-(2n-1)\over 4n^2-1}\cdot{1\over n!}\\
&=&\sum_{n=0}^{\infty}{2n^2+n\over 4n^2-1}\cdot{1\over n!}-\sum_{n=0}^{\infty}{2n-1\over 4n^2-1}\cdot{1\over n!}\\
&=&\sum_{n=1}^{\infty}{1\over 2n-1}\cdot{1\over (n-1)!}-\sum_{n=0}^{\infty}{1\over 2n+1}\cdot{1\over n!}
\end{eqnarray}
It is easy to check that the first and second series are the same and you can do the rest.
A: This is as simple as $1-2-3$.

$(1)$ 

Note that 
$$\begin{align}
\sum_{n=0}^\infty \frac{1}{(2n-1)n!}&=-1+\sum_{n=0}^\infty\frac{1}{(2n+1)(n+1)!}\\\\
&=-1+\sum_{n=0}^\infty \left(\frac{2}{2n+1}-\frac{1}{n+1}\right)\frac{1}{n!}\\\\
&=-1+\sum_{n=0}^\infty \left(\frac{2}{2n+1}\right)\frac1{n!}-\sum_{n=0}^\infty\left(\frac{1}{n+1}\right)\frac{1}{n!}\\\\
\end{align}$$


$(2)$

Hence, we see that
$$\begin{align}
\sum_{n=0}^\infty \left(\frac{1}{(2n-1)}-\frac{2}{2n+1}\right)\frac{1}{n!}&=-1-\sum_{n=0}^\infty\frac{1}{(n+1)!}\\\\
&=-1-(e-1)\\\\
&=-e
\end{align}$$


$(3)$

Finally, we have
$$\begin{align}
\sum_{n=0}^\infty \left(\frac12+\frac{1}{2(2n-1)}-\frac{1}{2n+1}\right)\frac{1}{n!}&=\frac12\sum_{n=0}^\infty \frac1{n!}+\frac12\left(\sum_{n=0}^\infty \left(\frac{1}{(2n-1)}-\frac{2}{2n+1}\right)\frac{1}{n!}\right)\\\\
&=\frac12 e-\frac12 e\\\\
&=0
\end{align}$$
as was to be shown!
A: $$\sum_{n\geq 0}\frac{1}{(2n-1)n!}=-1+\int_{0}^{1}\sum_{n\geq 1}\frac{x^{2n-2}}{n!} =-1+\int_{0}^{1}\frac{e^{x^2}-1}{x^2}\,dx$$
$$ \sum_{n\geq 0}\frac{2}{(2n+1)n!}=2\int_{0}^{1}e^{x^2}\,dx $$
and due to integration by parts:
$$ \int_{0}^{1}\frac{e^{x^2}-1}{x^2}\,dx = \left.-\frac{e^{x^2}-1}{x}\right|_{0}^{1}+2\int_{0}^{1}e^{x^2}\,dx $$
proving your $(3)$, then $(2)=S=0$.
