We say that a discrete random variable X has a geometric distribution with parameter $\alpha$ (where 0 < $\alpha$ < 1), if $\mathrm{Prob}\{X = n\} = P (n) = (1 − \alpha)\alpha^{n}, n = 0, 1, ...$ derive the formula for $\mathrm{Prob}\{X \geq n\}, n=0,1,...$
$\mathrm{Prob}\{X \geq n\} = \sum\limits_{n = 0}^\infty \mathrm{Prob}\{X=n\}=\sum\limits_{n = 0}^\infty (1-\alpha)\alpha^{n}= (1-\alpha)\sum\limits_{n = 0}^\infty \alpha^{n}=(1-\alpha)\dfrac{1}{1-\alpha}=1$
I believe I did something wrong here, can someone point the error and show correct solution?
First, since $n$ is fixed, we cannot use it as an index of summation. Write $\{X\geqslant n\}$ as the disjoint union of set $\{X=k\}$, $k\geqslant n$ in order to derive that $\mathbb P(X\geqslant n)=\sum_{k=n}^{\infty}\mathbb P(X=k)$. Then replace $\mathbb P(X=k)$ by its expression and do the change of index $j=k-n$.