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Show that $p(n)=\frac1{n(n+1)}$ for $n = 1, 2,...$ is a probability function for some discrete random variable $X$. Find $E(X)$.

So I know that $E(X)$ for a discrete random variable is $E(X)=x1p1+x2p2+...$ if $X$ is a discrete random variable that takes values $x1,x2,...$ with corresponding probabilities $p1,p2,..$. For this question is $p(n)$ the probability for any value of $n$? For example $p(1)=\frac1{2}$ and this would be the probability for x1? If this is true how would you go about proving that p(n) is a probability function? I assume once this is proven I can use the above equation to find $E(X)$? Am I close?

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So your RV is countable, i.e. defined on $\mathbb{Z}^{+}$. To prove that this is a pdf, show that (following Robert Israel's idea) $$ \lim_{n \to \infty}\bigg(1-\frac{1}{n+1}\bigg)=1 $$ To find the expectation, note that $$ \mathbf{E}X= \lim_{n \to \infty} \sum_{k=1}^{n}\frac{k}{k(k+1)}= \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k+1}= \lim_{n \to \infty} \bigg(H_{n+1}-1 \bigg)= \infty $$ where $H_n$ is nth Harmonic number

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@Alex-I dont understand the significance of the Harmonic number. Could you elaborate please? –  Sprock Oct 24 '12 at 13:23
    
$H_n =\sum_{k=1}^{n} \frac{1}{k}=O(\log n)$, so for $n \to \infty$ it diverges –  Alex Oct 24 '12 at 18:52
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Yes, $p(n)$ gives the probability of the value $n$ for your random variable. The requirements for a probability function are: all $p(n) \ge 0$ and the sum of $p(n)$ over all values of $n$ is $1$. So in this case you want $$ \sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$$ Hint: $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$.

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