Triple Summation?

$\displaystyle \prod _{k=1}^{j} (2k-1)=(2j-1)!!$

Also $\large{\frac{(2j)!}{(2j-1)!!}=2^{j}j!}$

Now we get

$\large{\displaystyle \sum^{i}_{j=1} (2j+1)2^{j}j!}$

$\large{\displaystyle \sum^{i}_{j=1} 2^{j+1}(j+1-1)j! +2^j j! }$

$\large{\displaystyle \sum^{i}_{j=1} 2^{j+1} (j+1)!-2^j j!}$

$\rightarrow 2^{i+1}(i+1)!-2$

Finally we have

$\large{\displaystyle \lim_{ h \to \infty}\displaystyle \sum^{h}_{i=1} \frac{2^{i+1}(i+1)!-2+2}{((i+1)!)^2}}$

$\large{\displaystyle \lim_{ h \to \infty}\displaystyle \sum^{h}_{i=1} \frac{2^{i+1}}{(i+1)!}}$

$\large{\displaystyle \sum_{k=2}^{\infty} \frac{2^k}{k!} \rightarrow \displaystyle \sum_{k=0}^{\infty} \frac{2^k}{k!}-\frac{1}{0!}-\frac{2}{1!} }$

We know that $\large{\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}=e^x}$

Hence

$\large{\boxed{e^2-3}}$