Why the equation can be equal to $(1-p)^k$?

Question

I was studying for geometric random variable, and I saw that

$$P(X>k)\sum_{i\ge k+1}p(1-p)^{i-1}$ =$(1-p)^k$$

I don't understand why it can be equal to $(1-p)^k$?

Answer 1

For simplicity, let's denote $1 - p$ by $q$ for a moment, assume $0 < q < 1$.

First, notice that \begin{align} & \sum_{i = k + 1}^\infty q^{i - 1} \\ = & q^k + q^{k + 1} + \cdots \quad \text{collecting the common term } q^k\\ = & q^k(1 + q + q^2 + \cdots) \\ = & q^k \frac{1}{1 - q} \\ = & \frac{q^k}{p} \end{align} where we used the celebrated geometric series summation formula: $$\boxed{1 + q + q^2 + \cdots = \frac{1}{1 - q}, \quad |q| < 1.}$$ Therefore the original summation becomes $$p \sum_{i = k + 1}^\infty q^{i - 1} = p \frac{q^k}{p} = q^k = (1 - p)^k.$$

If $p = 1$, the equality is trivial to verify (the equality does not hold for $p = 0$ though).

Answer 2

$\displaystyle \sum_{i\ge k+1}p(1-p)^{i-1} = \sum_{i+k+1\ge k+1}p(1-p)^{i+k}=\sum_{i\ge 0}p(1-p)^{i+k} = p(1-p)^{k}\sum_{i\ge 0}(1-p)^i$

And this is equal to $\displaystyle p(1-p)^k \times \frac{1}{1-(1-p)} = (1-p)^k$.