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Could someone help me understand how we get from sum to one in this formula of geometric distribution. Probably, some simplification I don't see

Formula

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The formula for the geometric distribution is $P(X=n)=p(1-p)^{n-1}$ with $n\in \mathbb{N}-\{0\}$, adding the terms we have:

$S=\sum_{n=1}^{\infty}P(X=n)=\sum_{n=1}^{\infty}p(1-p)^{n-1}$, we can change the variable like $j=n-1$, so if $n=1$ then $j=1-1=0$, then the sum $S$ is:

$S=\sum_{j=0}^{\infty}p(1-p)^{j}=p\sum_{j=0}^{\infty}(1-p)^{j}$

Using the geometric sum for $|x|<1$, $\sum_{j=0}^{\infty}x^j=\frac{1}{1-x}$, then S is equal to $S=p\frac{1}{1-(1-p)}=1$, where we used the geometric sum with $x=1-p$.

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Let $q:=1-p$. Then write the sum

$$S:=1+q+q^2+\cdots$$

You observe that

$$qS=q+q^2+q^3+\cdots=S-1$$ or $$S=\frac1{1-q}=\frac1p$$

so that

$$pS=1.$$

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