infinite fraction $\left(2e^z\right)$

This fraction can be expressed as: $1+\dfrac{3}{2}+\dfrac{5}{2\times4}+\dfrac{7}{2\times4\times6}+\dfrac{9}{2\times4\times6\times8}+\cdots=\dfrac{1}{2^0\times0!}+\dfrac{3}{2^1\times1!}+\dfrac{5}{2^2\times2!}+\dfrac{7}{2^3\times3!}+\dfrac{9}{2^4\times4!}+\cdots$ We change it into the summation form:

$\quad\displaystyle\sum_{k=0}^\infty \dfrac{2k+1}{2^kk!} \\ =\displaystyle\sum_{k=0}^\infty \dfrac{2k}{2^kk!}+\displaystyle\sum_{k=0}^\infty \dfrac{1}{2^kk!} \\= \displaystyle\sum_{k=1}^\infty \dfrac{1}{2^{k-1}(k-1)!} + \displaystyle\sum_{k=0}^\infty \dfrac{1}{2^kk!} \\=\displaystyle\sum_{k=0}^\infty \dfrac{1}{2^kk!} +\displaystyle\sum_{k=0}^\infty \dfrac{1}{2^kk!} \\=2\displaystyle\sum_{k=0}^\infty \dfrac{1}{2^kk!} \\=2\displaystyle\sum_{k=0}^\infty \left(\dfrac{1}{2}\right)^k\dfrac{1}{k!} \\= 2e^\frac{1}{2}=2\sqrt{e}\approx\boxed{3.29744}$

Let the given fraction be $Q$ . Then we have

$\begin{aligned} Q & = 1 + \frac 32 + \frac 5{2\cdot 4} + \frac 7{2\cdot 4 \cdot 6} + \cdots \\ & = \frac 1{2^0\cdot 0!} + \frac 3{2^1\cdot 1!} + \frac 5{2^2\cdot 2!} + \frac 7{2^3\cdot 3!} + \\ & = \sum_{n=0}^\infty \frac {2n+1}{2^nn!} \\ & = \sum_{n=1}^\infty \frac n{2^{n-1}n!} + \sum_{n=0}^\infty \frac 1{2^nn!} & \small \blue{\text{Replace }\frac 12 \text{ by }x} \\ & = \sum_{n=1}^\infty \frac {nx^{n-1}}{n!} + \sum_{n=0}^\infty \frac {x^n}{n!} \\ & = \frac d{dx}e^x + e^x \\ & = e^x + e^x & \small \blue{\text{Put back }\frac 12 \text{ as }x} \\ & = 2\sqrt e \approx \boxed{3.29744} \end{aligned}$