Respuesta :
Answer:
[tex]\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}[/tex]
Step-by-step explanation:
The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"
[tex]P(X=x)=(1-p)^{x-1} p[/tex]
Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:
[tex]X\sim Geo (1-p)[/tex]
In order to find the expected value E(1/X) we need to find this sum:
[tex]E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}[/tex]
Lets consider the following series:
[tex]\sum_{k=1}^{\infty} b^{k-1}[/tex]
And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:
[tex]\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}[/tex] (a)
On the last step we assume that [tex]0\leq r\leq b[/tex] and [tex]\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}[/tex], then the integral on the left part of equation (a) would be 1. And we have:
[tex]\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)[/tex]
And for the next step we have:
[tex]\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}[/tex]
And with this we have the requiered proof.
And since [tex]b=1-p <1[/tex] we have that:
[tex]\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}[/tex]
