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(I am confused here with the limits. It says x = 0,1,2,3... So what is my end limit her? Thanks.)

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up vote 2 down vote accepted

$\sum_{x=0}^\infty k(\frac{4}{5})^x = 1$

We are finding sum of a geometric series

A geometric series is of the form

$S_n = a + ar^1 + ar^2 + ... + ar^n$

$S_nr - S_n = ar^{n+1}- a$

$S_n = a[\frac{r^{n+1}-1}{r - 1}]$

In this case our r =4/5 and a= k

$1 = \lim_{n->\infty} k[\frac{(0.8^{n+1}) - 1}{0.8 - 1}]$

$1 = k \frac{0 - 1}{-0.2}$

$1 = k (5)$

$k = 1/5$

I am confused here with the limits. It says x = 0,1,2,3... So what is my end limit her? Thanks

Infinity, it keeps going

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This probability density function has mass on $\mathbb{N}_0$ which is all natural numbers including $0$. So you're looking at a random variable having mass at infinitely (though countably) many points, and hence this is an exercise in evaluating a series.

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That I know. But like, even if I substitute the values for x = 0,1,2,3, and then equate the whole thing to 1, i get the wrong answer for k. Because of the dots, so I dont knw the end limit. – har00n86 Aug 13 '13 at 11:23
Well, if you've already calculated $k$, then please add that to your question so that we a chance of knowing where you're stuck. Hint: It is not enough to sum the values for $x=0,1,2,3$, instead you have to sum over all natural numbers. This is what the dots in $0,1,2,3,\ldots$ is indicating. – Stefan Hansen Aug 13 '13 at 11:31
Can you show me how to add for all natural numbers? thanks – har00n86 Aug 13 '13 at 11:52

There is no end limit, it is just infinity... So, you have to sum up $g(x)$ from $x=0$ to $\infty$, and make it equal to $1$ to find $k$.

This can be easily done, since the summation is a geometric series:

$\sum_{x=0}^\infty k\left(\frac{4}{5}\right)^x = k \frac{1}{1-\frac{4}{5}} = 5k$.

So, $k$ should be equal to $\frac{1}{5}$.

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