Question: D = { mn/(1+m+n) } for m,n natural numbers.

To simplify the expression, I presumed m=n, which means:  

D = { n^2 / (2n + 1) }

Now, I know by intuition **there is no supremum**, for this series convergencing to infinity, but I couldn't find a formal way to prove it.

I thought about proving by contradiction:  

- assume S is the supremum  

- ϵ > 0  

S - ϵ > n^2 / (2n + 1) 

But then I got stuck.. 

Any guideness will be very appreciated.  

Thanks!

Answer: 
Hint: we can set $n = k$, $m = k-1$ to get 
$$
\frac{mn}{1+m+n}=
\frac{k(k-1)}{1 + k + k-1} = \frac{k(k-1)}{2k} = \frac{k-1}{2}
$$
Thus, $\frac{k-1}{2}\in D$ for every $k\geq 2$.
So, if $D$ has a supremum, then the supremum of $D$ is bigger than $\frac{k-1}{2}$ for every integer $k$. Why does this result in a contradiction?
**Hint:**
$$\frac{n^2}{2n+1} \ge \frac{n^2}{3n} =n/3$$