Probability interpretation

You have $P(X>4)=P(X=5)+...+P(X=n)+...=p\cdot\sum_{i=4}^{\infty}(1-p)^i$

This is a geometric series $a_1=(1-p)^4$ and $q=1-p$

By Geometric series formula, we have when $q<1$:

$lim_{n->\infty}\sum_{i=1}^{n}a_1 \cdot q^i = \frac{a_1}{1-q}=\frac{(1-p)^4}{1-(1-p)} = \frac{(1-p)^4}{p}$

But we had the multiplication with $p$ outside so we get $(1-p)^4$

It will work with any Geometric distribution, and for every value.

If you look for intuition, $(1-p)^4$ is the chance to "fail" in the first 4 times, so it leads to $P(X>4)$

Let $r\neq 1$. Then \begin{align*} \sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}, \end{align*} which for $r \in (0,1)$ tends to $\frac{1}{1-r}$ as $n$ tends to infinity.

Thus, if $p\in(0,1)$, then \begin{align*} p \sum_{k=4}^\infty (1-p)^k &= \left( p \sum_{k=0}^\infty (1-p)^k \right) - \left(p \sum_{k=0}^{4-1} (1-p)^k\right) \\ &= p\frac{1}{1-(1-p)} - p\frac{1-(1-p)^4}{1-(1-p)} \\ &= 1 - \left(1 - (1-p)^4\right) \\ &= (1-p)^4. \end{align*}

(In the above, the LHS of course corresponds to $\mathbb{P}(X>4) = \sum_{k=5}^\infty \mathbb{P}(X = k)$.)

Alternatively, one can re-write the LHS according to \begin{align*} p \sum_{k=4}^\infty (1-p)^k &= p \sum_{k=0}^\infty (1-p)^{k+4} \\ &= (1-p)^4 p \sum_{k=0}^\infty (1-p)^k, \end{align*} when the result is established by noting that the last term equals $\frac{1}{1-(1-p)} = \frac{1}{p}$.