**Unformatted text preview: **k =0 ∞ X l = k +1 P ( X = l )(2 k + 1) . We can exchange the series, since P ( X = l )(2 k + 1) ≥ 0, and it follows that ∞ X k =0 P ( X > k )(2 k + 1) = ∞ X k =0 ∞ X l = k +1 P ( X = l )(2 k + 1) = ∞ X l =1 l-1 X k =0 P ( X = l )(2 k + 1) = ∞ X l =1 P ( X = l ) l-1 X k =0 (2 k + 1) = ∞ X l =1 P ( X = l ) " 2 l-1 X k =0 k + l-1 X k =0 1 # = ∞ X l =1 P ( X = l ) " 2 l-1 X k =1 k + l-1 X k =0 1 # = ∞ X l =1 P ( X = l ) 2 l ( l-1) 2 + l = ∞ X l =1 P ( X = l ) l 2 = ∞ X l =1 l 2 P ( X = l ) = ∞ X l =0 l 2 P ( X = l ) = E [ X 2 ] . The fifth line was obtained using the expression for the sum of n integers, n X k =1 k = n ( n + 1) 2 . The given equality is confirmed. 2 3...

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- Spring '08
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