User:Matt Kwan/Sandbox

${\begin{aligned}\left(1+{\frac {r}{n}}\right)^{n}&=\sum _{k=0}^{n}{n \choose k}{\frac {r^{k}}{n^{k}}}\\&=\sum _{k=0}^{n}{\frac {n!r^{k}}{n^{k}k!\left(n-k\right)!}}\\&=1+\sum _{k=1}^{n}{\frac {n(n-1)\cdots (n-k+1)(n-k)!r^{k}}{n^{k}k!\left(n-k\right)!}}\\&=1+\sum _{k=1}^{n}{\frac {n(n-1)\cdots (n-k+1)r^{k}}{n^{k}k!}}\\&\leq 1+\sum _{k=1}^{n}{\frac {r^{k}}{k!}}\\\end{aligned}}$

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:

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

$\left({\frac {r}{2}}\right)^{n}$ 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):

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

therefore:

${\begin{aligned}\left(1+{\frac {r}{n}}\right)^{n}&\leq 1+\sum _{k=1}^{n}r^{\left\lceil r\right\rceil }\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{k-\left\lceil r\right\rceil }\\&=1+r^{\left\lceil r\right\rceil }\sum _{k=1}^{n}\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{k-\left\lceil r\right\rceil }\end{aligned}}$

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

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

which is a bound independent of $n$