How to prove that $0<e-(1+\frac{1}{n})^n<\frac{3}{n}$ I proved the first inequality $0<e-(1+\frac{1}{n})^n$ but now struggle with second part. I tried mathematical induction: 1. For $n=1$ $e-2<\frac{3}{1}$ 2. $e-(1+\frac{1}{n})^n<\frac{3}{n} \Rightarrow e-(1+\frac{1}{n+1})^{n+1}<\frac{3}{n+1}$

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 $e<3$).