# How to prove that $a_n =(1+\frac1n)^n$ for all $n\ge 1$ applies to $a_n=\sum_{k=0}^n \frac1{k!}(1-\frac1n)(1-\frac2n)\cdots(1-\frac{k-1}n)$?

In an assignment for the course Real-Analysis I have to show that for all $$n \ge 1$$ applies: $$a_n = \displaystyle \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right)$$ where $$a_n$$ is given as: $$a_n =\left(1+\frac1n\right)^n$$

I have to use the binomial formula for this: $$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{k} y^{n-k}$$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

• $\dfrac{1}{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right)=\color{Green}{\dfrac{?}{k!}}\dfrac{1}{n^{k-1}}$ and write this green color quantity as a factorial. Mar 8, 2021 at 21:45
• Hint $\frac {n-1}n=1-\frac 1n$ Mar 8, 2021 at 21:46

$$LHS = \sum_{k=0}^n \frac{1}{k!} \prod_{j=0}^{k-1}\left(1-\frac{j}{n} \right)$$

$$= \sum_{k=0}^n \frac{1}{k!} \frac{1}{n^{k-1}} \prod_{j=0}^{k-1}\left(n- j \right)$$

$$= \sum_{k=0}^n \frac{1}{n^{k-1}} \left(\frac{1}{k!}\prod_{j=0}^{k-1}\left(n- j \right) \right)$$

Note that $$\frac{1}{k!}\prod_{j=0}^{k-1}\left(n- j \right) = \binom {n}{k}$$

Hence $$LHS = \sum_{k=0}^n \binom {n}{k} \frac{1}{n^{k-1}}$$

$$= \left(1+\frac{1}{n} \right)^n$$

• A little mistake you made : $k!\neq\prod\limits_{j=1}^{k-1}{k!}$. Mar 8, 2021 at 22:46
• The index range needs $0$: $$\frac1{k!}\prod_{j=\color{#C00}{0}}^{k-1}(n-j)=\binom{n}{k}$$
– robjohn
Mar 9, 2021 at 3:50
• Fixed. Thanks a lot, CHAMSI, robjohn
– PTDS
Mar 9, 2021 at 4:41