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f(x) = 1/x for x = 1,2,3,...,n

Hint: The sum of the first positive n integers is n(n + 1)/2, and the sum of their squares is n(n + 1)(2n + 1)/6

I know mu/mean will be the sum of products of x and its probability of occurring over all x (through n in this case). I just don't know how to incorporate the formulas given in the hint into the general formulas for mean and variance:

[tex]\Sigma[/tex]xf(x) is the mean

[tex]\Sigma[/tex] (x - [tex]\mu[/tex])

^{2}f(x) is the variance.

Thank you for your help, (I don't know why mu is elevated.)

Jeff