Upper/lower bound for $e$: $\left (\frac{n+1}{n}\right )^n\leq e$ using $e^n\ge 1+n$

Question

Provided the fact $e^n\geq 1+n$ for every $n\in\mathbb{N}$, prove the lower bound $$\left (\frac{n+1}{n}\right )^n\leq e,\ \forall n\in\mathbb{N} $$

from analysis we know, this is true.

Very first attempt fails miserably: $$(n+1)^n\overset{?}\leq n^ne\Longrightarrow n+1\leq n\sqrt[n]{e} $$ provided fact is too weak to make any further progress on that attempt.

Next up: Binomial expansion $$\left (1+\frac{1}{n}\right )^n = \frac{1}{n^n}+\binom{n}{1}\frac{1}{n^{n-1}} +\ldots +\binom{n}{1}\frac{1}{n}+1\overset{?}\leq e $$ This is how it's done in analysis, but I can't see how we make use of provided fact.

Hints would be most welcome on how to proceed.

Answer 1

$$ e^n \geq (n+1) \Rightarrow n \geq \log(n+1) \Rightarrow 1 \geq \frac{1}{n} \log(n+1) = \log{(n+1)^\frac{1}{n}} $$ $$ m=\frac{1}{n} \Rightarrow 1 \geq \log{(\frac{1}{m}+1)^m} \Rightarrow e \geq (\frac{1+m}{m})^m $$

Answer 2

Well, I think that the binomial way is OK. I'd recommend trying $$ \binom n k \leq n^k. $$ This can be seen easily with expanding the binomial coefficient: $$ \binom n k = \frac{n(n-1)(n-2) \ldots (n-k+1)}{k!}.$$ Each of the parentheses is $\leq n$, there is $k$ of them, and the whole thing is divided by something $\geq 1$, so it should hold (at least if $n, k \leq 1$ and $0 \leq k \leq n$ — i. e. if the "normal" conditions for the coefficient hold).

From here you should be able to carry on yourself. If not, however, tell me and I can give more clues.

Answer 3

Fact: $e^x\geq 1+x$ for all $x\in\mathbb{R}$ For lower bound, we take $x=\frac{1}{n},n\in\mathbb{N}$ and necessary result follows. We can get more than that, though. Pick $x=-\frac{1}{n+1},n\in\mathbb{N}$: $$-\ln (1-\frac{1}{n+1})\geq \frac{1}{n+1}\iff \left (1+\frac{1}{n}\right )^{n+1}\geq e $$