1. ## Induction again

So, here I'm asking for a check..

For each natural number $n$ with $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.

Proof:

Basis $n=3$: true.
Assume for some $k$ with $k\geq3$, that $\left(1+\frac1{k}\right)^k<k$. Multiplying by $\left(1+\frac1{k}\right)$ yields $\left(1+\frac1{k}\right)^{k+1}<k+1$. But since $k+1>k\implies{1/(k+1)<1/k}\implies{}\left[1+\frac1{k+1}\right]^{k+1}<\left(1+\frac1{k}\right)^{k+1}<k+1$ as desired.
the set $T$ of natural numbers which make the above statement true is such that $3\in{}T$, and if $k\in{}T,~k\geq3$, then $k+1\in{}T$.
Therefore, by the principle of induction, for every natural number $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.

The part I worry about is when I show that k<k+1... Is this ok?

2. ## Re: Induction again

I pretty much agree with your proof. Your question reduces to: For any positive integer k, the function $f(x)=x^k$ is strictly increasing for $x>0$. The easy reason that this is true is that the derivative of f is positive. Alternatively, it's easy to give an inductive proof.

My only criticism is that the base case n = 3 is not really proved. I would add some some argument.

3. ## Re: Induction again

Originally Posted by VonNemo19
So, here I'm asking for a check..
For each natural number $n$ with $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.
Let
\begin{align*}e_n=\left(1+\frac{1}{n}\right)^n&=1 +\frac{n}{1!}\cdot\frac{1}{n}+\frac{n(n-1)}{2!}\cdot\frac{1}{n^2}+\cdots+\frac{1}{n^n}\;\; \;\text{binomial theorem}\\&<1+\frac{1}{1!}+\frac{1}{2!}\cdots+ \frac {1} {n!}\\&\le 1+\left(1+\frac{1}{2}+\frac{1}{2^2}\cdots+ \frac{1}{2^{n-1}} \right)\\&<1+2=3 \end{align*}

That has nothing to do with induction. It is a standard theorem in advanced calculus.
Moreover, more importantly, each term of the $e\text{-sequence}$ is $<3<n$