limit with summation and product 
Given  $L=\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\bigg(\frac{1}{r!}\prod^{r}_{i=1}\left(\frac{i}{2}+\frac{1}{3}\right)\bigg)$. then $\lfloor L \rfloor$ is 

Try: $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1}{r!}\bigg[\left(\frac{1}{2}+\frac{1}{3}\right)\cdot \left(\frac{2}{2}+\frac{1}{3}\right)\cdots \cdots \left(\frac{r}{2}+\frac{1}{3}\right)\bigg]$$
could some help me to solve it, thanks
 A: For any $|a| < 1$,$$
\sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n (ak + b) = \sum_{n = 1}^\infty \frac{(-a)^n}{n!} \prod_{k = 1}^n \left( -\frac{b}{a} - k \right) = \sum_{n = 1}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n}.
$$
Note that $|a| < 1$, by the generalized binomial theorem,$$
(1 - a)^{-\frac{b}{a} - 1} = \sum_{n = 0}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n},
$$
thus$$
\sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n (ak + b) = \sum_{n = 0}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n} - 1 = (1 - a)^{-\frac{b}{a} - 1} - 1.
$$
In this question, take $\displaystyle a = \frac{1}{2}$ and $\displaystyle b = \frac{1}{3}$, then$$
L = \sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n \left(\frac{k}{2} + \frac{1}{3}\right) = 2^{\frac{5}{3}} - 1,
$$
and $[L] = 2$.
A: Comment too long,
so entered as an answer:
Wolfy also says that
$\sum_{n=1}^∞ \prod_{j=1}^n(1/2 + 1/(b j)) 
= 2^{1+2/b} - 1
$.
Wolfy also says that
(the first line is what it returns,
the remaining lines
are my attempt
to properly simplify)
$\begin{array}\\
\sum_{n=1}^∞ \prod_{j=1}^n(\frac1{a} + \frac1{b j}) 
&= -\dfrac{(\frac{a - 1}{a})^{-a/b} (a (\frac{a - 1}{a})^{a/b} -
 (\frac{a - 1}{a})^{a/b} - a)}{a - 1}\\
&= -\dfrac{a - 1 - a(\frac{a - 1}{a})^{-a/b}}{a - 1}\\
&= \dfrac{ a(\frac{a - 1}{a})^{-a/b}}{a - 1}-1\\
&= \dfrac{ a^{1+a/b}}{(a - 1)^{1+a/b}}-1\\
&= \left(\dfrac{ a}{a - 1}\right)^{1+a/b}-1\\
\end{array}
$
when $|a|>1$.
