0

I want to prove this simple fact:

$\frac{n}{n+1} \geq \frac{1}{2}$ for all $n\in \mathbb{N}$.

Would this suffice: Proof by induction:

Base case: let $n = 1$, we have the result.

Inductive step: assume that for some $k\in \mathbb{N}$ we have $\frac{k}{k+1}\geq \frac{1}{2}$, we want to show that $\frac{k+1}{k+2}\geq \frac{1}{2}$.

Since $\frac{k}{k+1}\geq \frac{1}{2}$ we have that $2k\geq k+1$ iff $k\geq 1$. Thus, $k\geq 1$ and then $2k+2\geq k+2$, which simplifies to $\frac{k+1}{k+2}\geq \frac{1}{2}$. We are able to divide by these expressions as they are always positive.

Euler88 ...
  • 2,090

2 Answers2

1

For $n=1$ we have $n/(n+1) = 1/2$.

Let $n \geq 1$. If $n/(n+1) \geq 1/2$, i.e. if $2n \geq n+1$, then $2(n+1) = 2n+2 \geq n+1+2 = n+3 > n+2$, i.e. $(n+1)/(n+2) > 1/2$, qed.

Yes
  • 20,719
1

W/O induction, for $n>0,$ $$\dfrac n{n+1}>\dfrac12\iff 2n>n+1\iff n>1$$