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Given $f(z)=p_0 \left (\frac 56\right)^z$ for $z = 0, 1, 2, 3,\cdots$

Find the value for $p_0$ which makes $f(z)$ a valid pmf.

I know the sum of the pmf values must add to one but I am not sure how to use the given information to answer the question. Any suggestions?

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    Can you sum a geometric series? As you correctly point out we must have $1=p_0\sum_{n=0}^{\infty} \left( \frac 56 \right)^n$ – 2017-02-15
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    I am still not sure how I would solve for p0 – 2017-02-15
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    If $1 = p_0 S$ then $p_0= 1/S$. You just need to evaluate $S$, which is $\sum_{n=0}^\infty (\tfrac 56)^n$. As lulu suggests, research "geometric series". – 2017-02-15

1 Answers 1

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$$1=p_0(1+\frac{5}{6}+\frac{25}{36}+....)$$

$$1=p_0+\frac{5}{6}p_0+....\tag{1}$$

Multiply both sides by $\frac{5}{6}$:

$$\frac{5}{6}=p_0(\frac{5}{6}+\frac{25}{36}+....)$$

$$\frac{5}{6}=\frac{5}{6}p_0+....\tag{2}$$

Subtract equation $2$ from $1$:

$$1-\frac{5}{6}=p_0$$

$$\frac{1}{6}=p_0$$