Show that if X is a geometric random variable with parameter p, thenE[1/X]= −p log(p)/(1−p)Hint: You will need to evaluate an expression of the formi=1➝[infinity]∑(ai/ i)To do so, writeai/ i=0➝a∫(xi−1) dx then interchange the sum and the integral.

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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]