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While reading a proof i came across this step which i could not understand. The chunk below is part of bigger expression, but in the interest of reducing noise i am just posting the sub expression if needed please let me know and i'll post the full expression.

$\left| \dfrac {1} {n\left( c+n-1\right) }\right| =\frac1{n^2}\left| 1-\dfrac {c-1} {n}+O\left( \dfrac {1} {n^{2}}\right) \right| $

What result is being used to accomplish this ?

Edit: $\left| \dfrac {\left( a+n-1\right) \left( b+n-1\right) } {n\left( c+n-1\right) }\right| = \left| 1+\dfrac {a-1} {n}\right| \left| 1+\dfrac {b-1} {n}\right|\left| 1-\dfrac {c-1} {n}+O\left( \dfrac {1} {n^{2}}\right) \right| $

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    @joriki Thanks mate.2012-03-06

1 Answers 1

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Divide the top by $n^2$, simplifying as per the right-hand side. Divide the bottom by $n^2$. We want to study the behaviour of $\frac{1}{1+\frac{c-1}{n}}$ for $n$ large. Temporarily, let $\frac{c-1}{n}=x$. Note that $\frac{1}{1+x}$ has the familiar power series expansion $\frac{1}{1+x}= 1-x+x^2-x^3+ x^4-x^5+\cdots.$

Thus $\dfrac{1}{1+x}=1-x +O(x^2)$, which is exactly what you need.

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    @Hardy: You are very welcome. There was a $t$ypo a$t$ $t$he end, now fixed.2012-03-02