Why is $a_{n} < \sum_{r=0}^{n} \frac{1}{r!}$?

Question

Guys I have done a lot of thinking but I cannot see why this is true. Book says In (2.8), each of the factors

$ 1-(\frac{i}{n})$

is less than 1, and so from (2.6) we deduce that

$a_{n} < \sum_{r=0}^{n} \frac{1}{r!}$.

$a_{n}=\sum_{r=0}^{n}\binom{n}{r}\frac{1}{n^{r}}$ (2.6)

$\binom{n}{r}\frac{1}{n^r}=\frac{n(n-1)...(n-r+1)}{r!n^r}=\frac{1}{r!}(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{r-1}{n})$ (2.8)

I have tried many methods but I cannot see why this is true. I still think

$a_{n} > \sum_{r=0}^{n}\frac{1}{r!}$

and

$(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{r-1}{n})$

makes it increase. I hope I will get rigorous proof about it. Please Help me deduce it since I have very little time left. Thanks for the help in advance!

Answer 1

It’s actually quite clear

$$\binom{n}{r}\frac{1}{n^r}=\frac{n(n-1)...(n-r+1)}{r!n^r}=\frac{1}{r!}\underbrace{(1-\frac{1}{n})}_{<1}\underbrace{(1-\frac{2}{n})}_{<1}\cdots\underbrace{(1-\frac{r-1}{n})}_{<1}<\frac{1}{r!}$$ Now sum them up to get $a_n$.

Answer 2

This is because each term of $a_n$ satisfies $$\binom nr \frac 1{n^r}=\frac{n!}{r!(n-r)!}\frac1{n^r}=\frac 1{r!}\frac{n(n-1)\dotsm(n-r+1)}{n^r}<\frac1{r!}\frac{n^r}{n^r}=\frac1{r!}.$$