Prove: $(1-\frac{1}{d+1})^d>\frac{1}{e}$ 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!
 A: 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
$$
\bbox[5px,border:2px solid #C0A000]{\left(1-\frac1{d+1}\right)^d\gt\frac1e}
$$
A: 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 $\frac{d^2}{dx^2}\log(1+x)=-\frac{1}{(1+x)^2}<0$.
A: 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 :)
A: Hint: for $0<h<1,$
$$ \ln (1-h) = -(h+h^2/2 + h^3/3 + \cdots) >  -(h+h^2 + h^3 + \cdots)$$
A: 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.
A: 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 :)
A: 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
$$
\mathrm{e}^{-1}<\mathrm{e}^{-1}\left(1+\frac{1}{d+1} +\text{O}\left(\frac{1}{d+1}\right)^2\right)
$$
A: 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
