I was reading a proof of the central limit theorem, where some taylor expansions were involved. Eventually, I got that $$\log M_x(t)=n\left[\left(\frac{t^2}{2n}+ \frac{t^3}{6n^{3/2}}+\dotsb\right)-1/2\left(\frac{t^2}{2n}+ \frac{t^3}{6n^{3/2}}+\dotsb \right)^2+\dotsb\right].$$ It follows without proof that $ \lim_{n \to \infty} \log M_x(t)= t^2/2$, but I do not understand why this holds. I guess this has something to do with the fact that we are working with 'exponentially-fast converging' power series. Any hints are appreciated.
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Are $x$ and $n$ supposed to be the same here? – Greg Martin Apr 04 '17 at 19:52
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X stands for the random variable under which we compute the moment generating function. ( just ignore it. has really no connexion. my question is why we can take the first term and say this is the limit) – Dan Leonte Apr 04 '17 at 20:01
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In that case, you can't really mean $\lim_{x\to\infty}$, can you? – Greg Martin Apr 04 '17 at 20:02
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@GregMartin Yes You are right. That is a typo. I am sorry for the confusion and thank you for your observation. I changed it – Dan Leonte Apr 04 '17 at 20:54
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It is not clear what all of the integers in the denominators are. What are they? – DanielWainfleet Apr 04 '17 at 21:21
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1Ok, getting closer. Do you agree that, if one is allowed to take the $\lim_{n\to\infty}$ term-by-term on the right-hand side, that the result follows? If so, are you asking why that is in fact allowed? – Greg Martin Apr 04 '17 at 21:29
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@GregMartin yes that is what I am asking. Does it have something to do with uniform convergence? – Dan Leonte Apr 05 '17 at 13:05
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Yes, probably something like the Dominated Convergence Theorem is involved. Hard to tell without knowing the exact series in question. – Greg Martin Apr 05 '17 at 16:56