Tag: e

For $n \in \mathbb{Z}_{>0}$ prove,
$\left( 1 + \frac{1}{n} \right)^n = 1 + \sum_{k=1}^n \left( \frac{1}{k!} \cdot \prod_{r=0}^{k-1} \left(1 - \frac{r}{n}\right) \right).$
For $n > 1$ , prove
$2 < \left( 1 + \frac{1}{n} \right)^n < 1 + \sum_{k=1}^n \frac{1}{k!} < 3.$

Proof. Recall the Binomial theorem,

$(a+b)^n = \sum_{k=0}^n \binom{n}{k} \left( \frac{1}{n} \right)^k.$

So, letting $a = \frac{1}{n}$ and $b = 1$ we have,

$\begin{align*} \left(1 + \frac{1}{n}\right)^n &= \sum_{k=0}^n \binom{n}{k} \left(\frac{1}{n} \right)^k \\ &= 1 + \sum_{k=1}^n \binom{n}{k} \left( \frac{1}{n} \right)^k \\ &= 1 + \sum_{k=1}^n \frac{n!}{k! (n-k)!} \left( \frac{1}{n} \right)^k &(\text{Def. of } \binom{n}{k})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left( \left( \frac{\prod_{r=1}^n r}{\prod_{r=1}^{n-k} r} \right) \cdot \left( \frac{1}{n} \right)^k \right) &(\text{Def. of } n!)\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left(\prod_{r=n-k+1}^n r \right) \left( \frac{1}{n} \right)^k &(\text{cancelling})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left(\prod_{r=0}^{k-1} (n-r) \right) \left( \frac{1}{n} \right)^k &(\text{reindexing prod.})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} (n-r) \right) \left( \prod_{r=0}^{k-1} \frac{1}{n} \right) &(\text{rewritting } (1/n)^k \text{ as prod})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} \left( 1 - \frac{r}{n} \right) \right). \qquad \blacksquare \end{align*}$
Proof. First, we prove the left inequality,

$2 < \left( 1 + \frac{1}{n} \right)^n.$

If $n > 1$ , then $n \geq 2$ , so using the Binomial theorem we have,

$\left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^n \binom{n}{k} \left( \frac{1}{n} \right)^k \ \implies \ \left( 1 + \frac{1}{n} \right)^n = 1 + n\left(\frac{1}{n} \right) + \sum_{k=2}^n \binom{n}{k} \left( \frac{1}{n} \right)^k > 2.$

Where we know the inequality is strict since there is at least one term (which is necessarily positive) in $\sum_{k=2}^n \binom{n}{k} \left( \frac{1}{n} \right)^k$ since $n \geq 2$ .
Next, we prove the middle inequality,

$\left( 1 + \frac{1}{n} \right)^n < 1 + \sum_{k=1}^n \frac{1}{k!}.$

By part (a) we know,

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

Further, for $n > 1$ we have $\left(1 - \frac{r}{n}\right) < 1$ for all $r = 0, \ldots, n-1$ . Thus, we know that,

$\prod_{r=0}^{k-1} \left(1 - \frac{r}{n} \right) < \prod_{r=0}^{k-1} 1 = 1.$

Hence,

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

for all $k$ . Therefore, we have established the second inequality,

$\left(1 + \frac{1}{n} \right)^n = 1+ \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} \left( 1 - \frac{r}{n} \right) \right) < 1 + \sum_{k=1}^n \frac{1}{k!}.$

Finally, we prove the right inequality,

$1 + \sum_{k=1}^n \frac{1}{k!} < 3.$

Here we expand the first few terms and use a previous result,

$\begin{align*} 1 + \sum_{k=1}^n \frac{1}{k!} &= 1 + 1 + \frac{1}{2} + \frac{1}{6} + \sum_{k=4}^n \frac{1}{k!} \\ &< \frac{8}{3} + \sum_{k=4}^n \frac{1}{2^k} & (2^k < k! \implies \frac{1}{2^k} > \frac{1}{k!})\\ &= \frac{8}{3} + \frac{1}{16} \left( \sum_{k=0}^{n-4} \frac{1}{2^k} \right) & (\text{reindexing and pulling out } \frac{1}{2^4}) \\ &= \frac{8}{3} + \frac{1}{16} \left( 2 - \frac{1}{2^{n-4}}\right) \\ &= \frac{8}{3} + \frac{1}{8} - \frac{1}{2^n} \\ &< 3 \end{align*}$

for all $n > 1$ . In the second to last line we used this result on the $k$ th powers of a real number $x$ (in this case $x = \frac{1}{2}$ ). This completes the proof for all of the inequalities requested $. \qquad \blacksquare$