Why is 1-n*ln((2n+1)/(2n-1))~-1/(12x^2)?

**isler** · October 24th 2010, 06:19 PM

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?

**Opalg** · October 25th 2010, 02:30 AM

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 $\frac{2n+1}{2n-1}$ as $\dfrac{1+\frac1{2n}}{1-\frac1{2n}}$ . Then

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

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

**isler** · October 25th 2010, 06:44 AM

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

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