User:Matt Kwan/Sandbox

$$ \begin{align} \left ( 1 + \frac r n \right )^n &= \sum_{k=0}^n {n \choose k} \frac {r^k} {n^k} \\ &=\sum_{k=0}^n \frac{n!r^k}{n^k k! \left ( n-k\right )!}\\ &=1+ \sum_{k=1}^n \frac{ n(n-1)\cdots(n-k+1)(n-k)!r^k }{n^k k! \left ( n-k\right )!}\\ &=1+ \sum_{k=1}^n \frac{ n(n-1)\cdots(n-k+1)r^k}{n^k k!}\\ &\le 1+ \sum_{k=1}^n \frac {r^k} {k!}\\ \end{align} $$

At this stage $$2^{k-1}$$ was shown to be less than or equal to $$k!$$ which is correct but if this is done the conclusion of the proof is:

$$\left ( 1 + \frac r n \right )^n \le 1+ r \frac {1-\left(\frac r 2 \right)^n} {1-\frac r 2}$$

$$\left(\frac r 2 \right)^n$$ will increase without bound if $$r>2$$, so this method does not prove the general case that $$\left ( 1 + \frac r n \right )^n$$ is bounded. (When $$r=2$$ there is a divide-by-zero problem as well.)

A possible although unwieldy solution is to replace k! with something other than $$2^{k-1}$$. I propose a complete solution as follows:

(using $$\left \lceil n \right \rceil$$ as the ceiling function):

$$ k! \ge \left \lceil r+1 \right \rceil^{k-\left \lceil r \right \rceil} $$

therefore:

$$ \begin{align} \left ( 1 + \frac r n \right )^n &\le 1+ \sum_{k=1}^n r^{\left \lceil r \right \rceil} \left ( \frac r {\left \lceil r+1 \right \rceil} \right )^{k-\left \lceil r \right \rceil}\\ &=1+ r^{\left \lceil r \right \rceil} \sum_{k=1}^n \left ( \frac r {\left \lceil r+1 \right \rceil} \right )^{k-\left \lceil r \right \rceil} \end{align} $$

$$\sum_{k=1}^n \left ( \frac r {\left \lceil r+1 \right \rceil} \right )^{k-\left \lceil r \right \rceil}$$ is a GP, so using $$S_n=\frac{T_0(1-r^n)}{1-r}$$ with $$n$$ as the number of terms and $$r$$ as the growth per term:

$$ \sum_{k=1}^n \left ( \frac r {\left \lceil r+1 \right \rceil} \right )^{k-\left \lceil r \right \rceil} = \frac {\left ( \frac r {\left \lceil r+1 \right \rceil} \right )^{1-\left \lceil r \right \rceil}\left(1-\left ( \frac r {\left \lceil r+1 \right \rceil} \right)^n \right)} {1- \frac r {\left \lceil r+1 \right \rceil}} $$

But for all $$r>0$$:

$$0<\frac r {\left \lceil r+1 \right \rceil} <1$$

So $$0<1-\left ( \frac r {\left \lceil r+1 \right \rceil} \right)^n<1$$

therefore:

$$ \left ( 1 + \frac r n \right )^n \le 1+\frac { r^{\left \lceil r \right \rceil} \left(\frac r {\left \lceil r+1 \right \rceil} \right)^{1-\left \lceil r \right \rceil}} {1- \frac r {\left \lceil r+1 \right \rceil}} $$

which is a bound independent of $$n$$