The back of my textbook says: 1-n*ln((2n+1)/(2n-1)) = O(1/n^2) Maple verified that: [1-n*ln((2n+1)/(2n-1))]/[-1/(12x^2)]->1 as n->infinity. But all my calculations show it to be order 1/n. What am I missing?

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Originally Posted by isler The back of my textbook says: 1-n*ln((2n+1)/(2n-1)) = O(1/n^2) Maple verified that: [1-n*ln((2n+1)/(2n-1))]/[-1/(12x^2)]->1 as n->infinity. But all my calculations show it to be order 1/n. What am I missing? In fact, the 1/n term turns out to be 0. Write as . Then So .

Thank you. I had been looking at it as ln(1+2/(2n-1)).

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