Verify that the following p.m.f. is valid and if it is, then find its c.d.f.
$$\mathbb{P}(X=n)=\frac{1}{n(n+1)}\;\forall\;n\in\mathbb{N}^+$$
My Attempt Verifying that it is valid is easy. I just need to check whether or not it sums to $1$. So $$\sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$ We know that this infinite series is true, so our p.m.f. is valid. Now to find the c.d.f.. Since this p.m.f. is discrete, I will use the following formula: $$F(t)=\sum_{k\in X, k\leq t}p(k)$$ My question is how should I divide the intervals so that I can write the c.d.f.? It is a piecewise function that changes with each $n$ so would my c.d.f. be \begin{equation} F(t)= \begin{cases} 0 & \text{if } t\leq0\\ \frac{1}{1(1+1)} &\text{if } 0<t\leq1\\ \frac{1}{2(2+1)} &\text{if } 1<t\leq2\\ ... \end{cases} \end{equation}
This doesn't seem like a very good way to write the c.d.f. so I am wondering if there are other ways to express it.
You can use induction as pointed out in the hint, or you can see it as follows:
$$ \mathbb{P}(X\leq n) = \sum_{i=1}^n \frac{1}{i(i+1)}=\sum_{i=1}^n \frac{(i+1)-i}{i(i+1)} = \sum_{i=1}^n \left(\frac{1}{i} -\frac{1}{i+1} \right)= 1-\frac{1}{n+1}. $$