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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- Oct 24th 2010, 06:19 PM #1

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- Oct 25th 2010, 02:30 AM #2
In fact, the 1/n term turns out to be 0. Write $\displaystyle \frac{2n+1}{2n-1}$ as $\displaystyle \dfrac{1+\frac1{2n}}{1-\frac1{2n}}$. Then

$\displaystyle \begin{aligned}\textstyle\ln\left(\frac{2n+1}{2n-1}\right) &= \textstyle\ln\bigl(1+\frac1{2n}\bigr) - \ln\bigl(1-\frac1{2n}\bigr) \\ &= \textstyle\frac1{2n} - \frac12\bigl(\frac1{2n}\bigr)^2 + O(n^{-3}) \ + \ \frac1{2n} + \frac12\bigl(\frac1{2n}\bigr)^2 + O(n^{-3}) \\ &= \textstyle\frac1n + O(n^{-3}).\end{aligned}$

So $\displaystyle 1-n\ln\left(\frac{2n+1}{2n-1}\right) = 1-n\bigl(\frac1n + O(n^{-3})\bigr) = O(n^{-2})$.

- Oct 25th 2010, 06:44 AM #3

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