Given that f(n)>0 and constant c>0 Prove that f(n)+c = O(f(n)) or provide a counter-example if it's false
This is false. A counterexample is $f(n)=1/n$ and $c=1$.

If $f(n)+c=O(f(n))$, then there exists $n_0&gt;0$ and $d&gt;0$, for any $n&gt;n_0$,  $f(n)+c\le df(n)$, or $1/n+1\le d/n$. But let $n=\max\{n_0+1,d\}$, the inequality fails obviously, a contradiction.

In fact, any $f(n)$ and $c$ such that $f(n)\to 0$ ($n\to\infty$) and $c&gt;0$ are sufficient.