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So I was working on a problem where I am provided with a PMF $p_X(k)= c/3^k$ for $k=1,2,3....$ I was able to calculate $c$ using the basic property of PMF and it came to be 2. I am not able to solve the next part which states that "Find $P(X\ge k)$ for all $k=1,2,3......$.

Any suggestions?

P.S :Here is the actual question:

Let X be a discrete random variable with probability mass function $p_X(k) = c/3^k$ for k = 1, 2, ... for some $c > 0$. Find $c$. Find $P(X\ge k)$ for all $k = 1, 2,3....$

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    Are you aware that $P(X\geq k)=\sum_{x\geq k} p_X(x)?$ – 2017-02-16
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    I was not aware of that. So then all I need to do is just put in the value of the PDF and evaluate the summation? – 2017-02-16
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    Yes, that's correct – 2017-02-16
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    Okay, I am stuck again. $x \ge k$ , how do I put that into the summation? I mean let $k=1$. So we have $x \ge 1$. Then how do I proceed from there? I mean k has infinite values, so how do I put them into the summation? – 2017-02-16
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    Geometric series. – 2017-02-16
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    That is true for x=1 But then again I have to put k=2 and then evaluate the whole summation again? – 2017-02-16
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    Use _general_ formula for geometric series. – 2017-02-16
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    $P(X\geq k) = 1-P(X – 2017-02-16

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The idea of Alex R. is almost right. $k$ starts at $1$ therefore

$$P(X\geq k)=1-P(X

$$\sum_{x=1}^{k-1}p_X(k)=c\sum_{x=1}^{k-1}\left( \frac{1}{3} \right)^x$$

The closed form of a geometric series is

$\sum_{x=1}^{k} a^x=a\cdot \frac{1-a^k}{1-a}$

Thus

$c\sum_{x=1}^{k-1}\left( \frac{1}{3} \right)^x=\frac{c}3\cdot \frac{1-\frac1{3^k}}{1-\frac13}-c\left( \frac13 \right)^k=\frac{c}3\cdot \frac{1-\frac1{3^k}}{\frac23}-c\left( \frac13 \right)^k=\frac{c}3\cdot \frac32\cdot (1-\frac1{3^k})-c\left( \frac13 \right)^k$

$P(X

The sum of all probabilities add up to 1.

$$\lim_{k \to \infty}\frac{c}2\cdot (1-\frac1{3^k})-c\left( \frac13 \right)^k=\frac{c}2=1\rightarrow c=2$$

$P(X

Consequently

$P(X\geq k)=1-\left( 1-\left( \frac13 \right)^{k-1} \right)=\left( \frac13 \right)^{k-1} $