Prove: $(1-\frac{1}{d+1})^d>\frac{1}{e}$

Question

I need to prove that $\left(1-\frac{1}{d+1}\right)^d>\frac{1}{e}$.

I guess that I have to use that $\left(1+\frac{1}{n+1}\right)^n\rightarrow e$ for $n\rightarrow\infty$ or better $<e$ or $\left(1+\frac{1}{n-1}\right)^n>e$ but I don't know exactly how it works.

Thank you for your help!

Answer 1

The given inequality is equivalent to: $$\left(1+\frac{1}{d}\right)^d < e $$ or to: $$\log\left(1+\frac{1}{d}\right)<\frac{1}{d}$$ that is trivial by concavity, since $\frac{d^2}{dx^2}\log(1+x)=-\frac{1}{(1+x)^2}<0$.

Answer 2

Hint: for $0<h<1,$

$$ \ln (1-h) = -(h+h^2/2 + h^3/3 + \cdots) > -(h+h^2 + h^3 + \cdots)$$

Answer 3

Well if you know l'hospitals rule you can show that $$(1-\frac{1}{x})^{x}$$ is equal to $e^{-1}$ as $x \to \infty$ and use this fact to prove your equation. If you don't, e can be defined as $$lim_{x->\infty}(1+\frac{1}{x})^{x}$$ and the function is less than that of e Use this facts plus $$lim_{x->\infty}(1+\frac{a}{x})^{x}=e^{a}$$ (try to prove this yourself as exercise) Substitute the $x$ for $d+1$ and the $a$ for -1 all this guides you to the answer

Answer 4

First, $$ (1 + \frac{1}{n})^n < e, $$ so $$ \frac{1}{(1 + \frac{1}{n})^n} > \frac{1}{e}. $$ Now, $$ (1 - \frac{1}{n + 1}) = \frac{n}{n + 1} = \frac{1}{\frac{n + 1}{n}} = \frac{1}{\left(1 + \frac{1}{n}\right)}, $$ so $$ (1 - \frac{1}{n + 1})^n = \frac{1}{(1 + \frac{1}{n})^n} > \frac{1}{e} $$ Q. E. D.

Answer 5

Clearly , $e^x > 1+x$ for x>0,

So $\frac{1}{1+x}>e^{-x}$.......(1)

Since, $(1-\frac{1}{(d+1)})$=$\frac{d}{d+1}$=$\frac{1}{1+\frac{1}{d}}>e^{-\frac{1}{d}}$ using...(1)

Thus, $(1-\frac{1}{(d+1)})$ > $e^{-\frac{1}{d}}$

And $(1-\frac{1}{(d+1)})^d$ > $e^{-1}$

Hope, this answers the question :)

Answer 6

my two pence(cents) worth.. $$ \left(1-\frac{1}{d+1}\right)^d = \frac{\left(1-\frac{1}{d+1}\right)^{d+1}}{1-\frac{1}{d+1}}\to \lim_{d>>1}\left(1-\frac{1}{d+1}\right)^{d+1}\left(1+\frac{1}{d+1} +\text{O}\left(\frac{1}{d+1}\right)^2\right)<\mathrm{e}^{-1}\left(1+\frac{1}{d+1} +\text{O}\left(\frac{1}{d+1}\right)^2\right) $$ and finally $$ \mathrm{e}^{-1}<\mathrm{e}^{-1}\left(1+\frac{1}{d+1} +\text{O}\left(\frac{1}{d+1}\right)^2\right) $$

Answer 7

Approach 1

Using Bernoulli's Inequality, which is strict for $d\ge2$, $$ \begin{align} \frac{\left(1-\frac1{d\vphantom{+1}}\right)^{d-1}}{\left(1-\frac1{d+1}\right)^d} &=\frac{\left(\frac{d-1}{d\vphantom{+1}}\right)^{d-1}}{\left(\frac{d}{d+1}\right)^d}\\ &=\frac{d}{d-1}\left(\frac{d^2-1}{d^2}\right)^d\\[9pt] &\gt\frac{d}{d-1}\left(1-\frac1d\right)\\[15pt] &=1 \end{align} $$ Thus, $\left(1-\frac1{d+1}\right)^d$ is strictly decreasing and its limit is $\frac1e$. Therefore, we have $$ \bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e} $$

---

Approach 2

If $x\ge-n$, Bernoulli's Inequality says that for $x\ne0$ and $n\ge2$, $$ 1+x\lt\left(1+\frac x2\right)^2\le\left(1+\frac xn\right)^n $$ Therefore, taking the limit as $n\to\infty$, we get that for $x\ne0$, $$ 1+x\lt e^x $$ Thus, $$ 1+\frac1d\lt e^{1/d} $$ Raise both sides to the $-d$ power $$ \bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e} $$

Answer 8

Why all so complicated? We have
$$\left(\frac{n+1}{n}\right)^n = \left(1 + \frac{1}{n}\right)^n < e $$ $$\Rightarrow \left(\frac{n}{n+1}\right)^n = \left(1 - \frac{1}{n+1}\right)^n > \frac{1}{e} $$ Done :)