Let $X,Y$ be discrete independent random variables with $X,Y \geq 0$. Then $$P(X+Y=n) = \underset{k=0}{\overset{n}{\sum}} P(X=n-k)P(Y=k).$$
(I am sure this is very very easy and I am just overlooking something very simple)
$\textbf{Proof:}$ Trying to show right hand side is the same as left hand side.
$$\sum_{k=0}^n P(X=n-k)P(Y=k) = \sum_{k=0}^n P(X=n-k,Y=k)$$ since $X,Y$ independent, (At this point I am just trying things to see if I can get the conclusion) $$ = \sum_{k=0}^n P(X+Y=n,Y=k)$$ $$ = \sum_{k=0}^n P(X+k=n)$$ $$ = P(X=n) + P(X=n-1)+ \cdots +P(X=0)$$ $$ = F_X(n)$$
Could anyone help with this?
*Thank you for your responding André and Dilip.
It is very clear now.