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

Question

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...

Answer 1

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.

Answer 2

This is as simple as $1-2-3$.

$(1)$

Note that

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

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$(2)$

Hence, we see that

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

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$(3)$

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!

Answer 3

$$\sum_{n\geq 0}\frac{1}{(2n-1)n!}=-1+\int_{0}^{1}\sum_{n\geq 1}\frac{x^{2n-2}}{n!} =-1+\int_{0}^{1}\frac{e^{x^2}-1}{x^2}\,dx$$ $$ \sum_{n\geq 0}\frac{2}{(2n+1)n!}=2\int_{0}^{1}e^{x^2}\,dx $$ 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 $(2)=S=0$.