# Understanding a simplification in a theorem

I'm trying to understand a theorem in a paper on page 14/24. We are given that

$$Z = (nq-1) \log \left(\frac{M+nq-1}{nq-1} \right) + M \log \left(\frac{M+nq-1}{M} \right) + \frac{1}{2} \log \left( \frac{M+nq-1}{(nq-1)M} \right) + O(1) .$$

Since $e^a > \left(1+\frac{a}{b} \right)^b$ for all $a > 0$ and positive integers $b$, the second term equals

$$\log \left(1 + \frac{nq-1}{M} \right)^M < \log e^{nq-1}$$

for all positive M and $nq-1 > 0$. This much I'm following, but then it goes on saying:

Since $0 < q < n$ and $n \leq M \leq qn^2$,

$$\dfrac{1}{2(nq-1)} \log \left( \frac{M+nq-1}{(nq-1)M} \right)\to 0 \quad \text{for } n \to \infty .$$

Where does the factor $\frac{1}{2(nq-1)}$ come from? This might even be trivial, but somehow I'm totally unable to see it.

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Could you provide a linkt to the paper? –  Raskolnikov Jun 7 '11 at 21:07
Given $\dfrac{1}{2(nq-1)} \log \dfrac{M+nq-1}{(nq-1)M}$, do you see why this tends to 0 as n tends to $\infty$? or do you also have a question about how to get this term? –  Edison Jun 7 '11 at 21:10
@El G It's about getting the term mostly. –  Pele Jun 7 '11 at 21:14
Does $\dfrac{1}{2(nq-1)} \log \dfrac{M+nq-1}{(nq-1)M} \to 0$, for $n \to \infty$ refer to the last term? –  Edison Jun 7 '11 at 21:18
Yes, apparently so. –  Pele Jun 7 '11 at 21:20

OK, thanks to the paper, I figured it out. What they do is rewrite

$$Z = (nq-1) \log \dfrac{M+nq-1}{nq-1} + M \log \dfrac{M+nq-1}{M} + \dfrac{1}{2} \log \dfrac{M+nq-1}{(nq-1)M} + O(1)$$

with their formula for the second term

$$M\log\left(1 + \dfrac{nq-1}{M}\right) < (nq-1)\log e$$

which gives

$$Z < (nq-1) \log \dfrac{M+nq-1}{nq-1} + (nq-1)\log e + \dfrac{1}{2} \log \dfrac{M+nq-1}{(nq-1)M} + O(1)$$

Of course, it has become an inequality, particularly an upper bound for $Z$. Now separate out the factor $(np-1)$ and you'll get

$$Z < (nq-1) \left( \log \dfrac{M+nq-1}{nq-1} + \log e + \dfrac{1}{2(nq-1)} \log \dfrac{M+nq-1}{(nq-1)M} + O\left(\frac{1}{nq-1}\right) \right)$$

Now, see the third term is exactly what they are computing the limit of. Since that goes to zero, and the $O\left(\frac{1}{nq-1}\right)$ term does as well, you get the end result

$$Z < (nq-1)\left( \log \dfrac{M+nq-1}{nq-1} + \log e \right) \; .$$

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Ahh, I see. Thanks a lot! –  Pele Jun 8 '11 at 9:28