Probability density,cumulative function problem

[atrus_ovis]

Homework Statement

Prob. density function is given below, representing probability of answer in a telephone service center
f(x) = c 2^x-1/ 5^x

where X is the random variable representing tries (phone calls).
a)find c
b)find the cumulative distribution function
i think i can solve the rest if i get through the first obstacles.

Homework Equations

The sum of a discrete density function from -inf to inf equals 1

The Attempt at a Solution

X is discrete, and the sum from -inf to inf must equal 1.
I sum up from 0 to inf (as negative phone calls don't make sense. is that right to do?) and come up with c=3/4
SUM c 2^x+1 /5^x , from 0 to inf
=lim, k->inf of SUM (1/3)^x , x from 0 to k
=((c/2) * 1-(1/3)^{inf + 1} )/(2/3) = 1
=>c=4/3

So f is now simplified to 1/3^x+1

Is the above method correct? Sadly, i then take the sum from 0 to inf to verify that it equals 1, but it doesn't.
How can i find the cumulative distribution?

 

[fzero]

The Attempt at a Solution

X is discrete, and the sum from -inf to inf must equal 1.
I sum up from 0 to inf (as negative phone calls don't make sense. is that right to do?) and come up with c=3/4
SUM c 2^x+1 /5^x , from 0 to inf
=lim, k->inf of SUM (1/3)^x , x from 0 to k
=((c/2) * 1-(1/3)^{inf + 1} )/(2/3) = 1
=>c=4/3

So f is now simplified to 1/3^x+1

You seem to have multiple mistakes here. The sum is

$$c \sum_{x=0}^\infty \frac{2^{x-1}}{5^x} = \frac{c}{2} \sum_{x=0}^\infty \left(\frac{2}{5}\right)^x$$

Also $$2/5\neq 1/3$$. You want to apply the formula for a geometric sum directly to the expression with 2/5 in it.

 

[atrus_ovis]

Ah, ok.
Then c = 6/5, and f(x) is simplified to (3/5 )* (2/5)^x