How do I prove $(1 + \frac{1}{n})^n < n$ by mathematical induction?

Question

$\displaystyle(1 + \frac{1}{n})^n < n$ for $n \gneq 3$

yes for $n = 1$ it is true

I assume it is true for $n = k$ and get

$\displaystyle(1+\frac{1}{k})^k < k$

I then go to $\displaystyle(1 +\frac{1}{k+1})^{k+1} < k+1$ and now I spend an hour doodling.

Answer 1

Consider $$\left(1 + \frac{1}{n+1}\right)^{n+1} < \left(1 + \frac{1}{n}\right)^{n+1} = \left(1 + \frac{1}{n}\right)^{n}\left(1 + \frac{1}{n}\right) < n\left(1 + \frac{1}{n}\right) = n+1$$ where the last inequality is due to the induction hypothesis.

Answer 2

First, the base case starts at $n=4$, not $n=1$, since it's not true for $n=1$. This is true for $n=4$ by direct calculation.

So assume $(1+\frac{1}{k})^k < k$. Then $$ \left(1+\dfrac{1}{k+1}\right)^{k+1} = \left(1+\dfrac{1}{k+1}\right)^k\left(1+\dfrac{1}{k+1}\right)< \left(1+\dfrac{1}{k}\right)^k\left(1+\dfrac{1}{k}\right) $$ since $1+\dfrac{1}{k+1}\lt 1+\dfrac{1}{k}$. Then apply your induction hypothesis.