How to prove $\frac{1}{4n} < e- \left(1 + \frac{1}{n} \right)^n$ Tried prove inequality
$$\frac{1}{4n}< e -\left(1 + \frac{1}{n}  \right)^n. $$
spent a lot of time. But i proved it through the Taylor series
$$ e^{\ln((1+\frac{ 1 }{ n } )^{n})}= e^{n\ln (1+\frac{ 1 }{ n })} = e^{1-\frac{ 1 }{ 2n } + o(\frac{ 1 }{ n^{2} } ) }. $$
Lector said that this was school lvl, and gave me hint
$$e = (1+\frac{ 1 }{ 2n })^{2n}$$
and
$$ (1+\frac{ 1 }{ 2n })^{2n} - (1+\frac{ 1 }{ n } )^{n} = (1+\frac{ 1 }{ n } )^{n}((1+\frac{ 1 }{ 2n })^{2n} (\frac{ n }{ n+1 })^{n} - 1 ).$$
I can't proove thath without Taylor series. Pls, halp me. I need to know how to do that.
Any solution will do, but without Taylor and L'Hopital
 A: I don't know if this is what you want. Let $x=\frac{1}{n}$ and then $x\in(0,1]$ and
$$ (1+x)^\frac1x\ge 2. $$
Define
$$ g(x)=e-(1+x)^{\frac1x}-\frac14x=e-e^{\frac1x\ln(1+x)}-\frac14x. $$
So
\begin{eqnarray}
g'(x)&=&-(1+x)^{\frac1x}\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x(1+x)}\right)-\frac14 \\
&=&(1+x)^{\frac1x-1}\frac{(1+x)\ln(1+x)-x}{x^2}-\frac14.
\end{eqnarray}
Define
$$ h(x)=(1+x)\ln(1+x)-x-\frac14x^2 $$
and then
$$ h'(x)=-\frac x2+\ln(1+x)>0, x\in[0,1]. $$
So $h(x)$ is increasing and hence $h(x)>h(0)>0$ for $x>0$. Therefore
$$ (1+x)\ln(1+x)-x\ge\frac14x^2 $$
and
$$ g'(x)= \frac14(1+x)^{\frac1x-1}-\frac14>\frac1{2(1+x)}-\frac14=\frac{1-x}{2(1+x)}>0 $$
which implies that $g(x)$ is increasing. Thus $g(x)>g(0^+)=0$ for $x>0$.
A: The hint was probably $e\gt\left(1+{1\over2n}\right)^{2n}$, not $e=\left(1+{1\over2n}\right)^{2n}$. With this, using $\left(1+{1\over2n}\right)^2=1+{1\over n}+{1\over4n^2}$ and the algebraic identity $a^n-b^n=(a-b)(a^{n-1}+\cdots+b^{n-1})$, where there are $n$ terms in the sum $a^{n-1}+\cdots+b^{n-1}$, one has
$$\begin{align}
e-\left(1+{1\over n}\right)^n
&\gt\left(1+{1\over2n}\right)^{2n}-\left(1+{1\over n}\right)^n\\
&=\left(\left(1+{1\over n}+{1\over4n^2}\right)-\left(1+{1\over n}\right)\right)\left(\left(1+{1\over n}+{1\over4n^2}\right)^{n-1}+\cdots+\left(1+{1\over n}\right)^{n-1}\right)\\
&\ge{1\over4n^2}(1+\cdots+1)\\
&={1\over4n}
\end{align}$$
