How to prove that $0I proved the first inequality $0<e-(1+\frac{1}{n})^n$ but now struggle with second part. I tried mathematical induction:


*

*For $n=1$  $e-2<\frac{3}{1}$

*$e-(1+\frac{1}{n})^n<\frac{3}{n} \Rightarrow e-(1+\frac{1}{n+1})^{n+1}<\frac{3}{n+1}$

 A: With analysis, without induction.
Define for $\;x>0\;$ :
$$f(x):=\left(1+\frac1x\right)^x+\frac3x-e\implies f'(x)=\left(1+\frac1x\right)^x\left(\log\left(1+\frac1x\right)-\frac1{x+1}\right)-\frac3{x^2}$$
But
$$g(x):=\log\left(1+\frac1x\right)-\frac1{x+1}\implies g'(x)=-\frac1{x(x+1)}+\frac1{(x+1)^2}=$$
$$=\frac1{x+1}\left(\frac1{x+1}-\frac1x\right)<0\;,\;\;\forall\,x>0\implies$$
$\;g(x)\;$ monotone desscending, and since $\;\begin{cases}g(x)\xrightarrow[x\to 0^+]{}\infty\\{}\\g(x)\xrightarrow[x\to\infty]{}0\end{cases}\;$ , we get that in fact $${}$$
$\;f'(x)=:g(x)>0\;,\;\;\forall\,x>0\;$ , which means $\;f(x)\;$ is monotone descending, and once more:
$$\begin{cases}f(x)\xrightarrow[x\to 0^+]{}\infty\\{}\\f(x)\xrightarrow[x\to\infty]{}0\end{cases}$$
we get that $\;f(x)>0\;$ forall $\;x>0\;$ ,which is what we wanted to prove
A: The inequalities
$$ \Big(1+\frac1n\Big)^n < e < \Big(1+\frac1n\Big)^{n+1} $$
are fairly standard.  You seem to know the first one; the second one has a similar proof (using GM/HM instead of AM/GM), and yields your upper inequality readily (if we also know that $e<3$).
