Finding the simplest form How can we simplify the expression below:
For any real numbers $a$ and $b$,
$
\frac{\frac{\frac{a+b}{2}+ b}{3}+ b} {4}+ b \cdots $
This is same as asking what is  $$\frac{a+ b\sum_{i=1}^ni!}{n!}$$ as $n$ approaches to $+ \infty$?
Thanks.
 A: Let's rewrite your sum like this:
$\dfrac{a+ b\sum_{i=1}^ni!}{n!}=\dfrac{a+ b \left( 1!+2!+ \cdots + \left( n-2\right)! + \left( n-1\right)! +n! \right)}{n!}$
$=\dfrac{a}{n!}+b \left( \dfrac{1!}{n!} +\dfrac{2!}{n!} + \cdots \dfrac{ \left( n-2\right)!}{n!} + \dfrac{\left( n-1\right)!}{n!} + \dfrac{n!}{n!}  \right)$
$=\dfrac{a}{n!}+b \left( \dfrac{1}{n!} +\dfrac{2}{n!} + \cdots \dfrac{ 1 }{n \left( n-1 \right)} + \dfrac{1}{n} + 1  \right)$
Take the limit as $n$ approches $+ \infty$ and you get $b$.
A: Because of $\sum_{i = 1}^n i = \frac{n (n+1)}{2}$ for all $n \in \mathbb{N}$ we have 
$\begin{align} \frac{a + b \sum_{i = 1}^n i}{n!} &= \frac{a}{n!} + \frac{b (n+1)}{2 (n-1)!} \\ &= \frac{a}{n!}+ \frac{b}{2(n-2)!} + \frac{b}{(n-1)!}
\end{align}$
for all $n \geq 2$ and each of these summands converges to $0$ for $n \rightarrow \infty$. 
Hence $\lim\limits_{n\rightarrow \infty} \frac{a + b \sum_{i = 1}^n i}{n!} = 0$ for all $a,b \in \mathbb{R}$.
EDIT:
As noted by lsp this does not answer the question.
