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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
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 $[L] = 2$.
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 $|a|>1$.
