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Suppose one knows that for a random function $f(n)$, $f(n)-a$ decays at some rate given by: $$Pr(|f(n)-t|>\epsilon)=g(\epsilon),$$for $g(\epsilon)\to0$, all as $n\to\infty$.

If the above holds, then plainly in addition to $f(n)\to a$, we have $1/f(x)\to 1/a$.

But at what rate?

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  • $\begingroup$ You need a sequence of RV's to talk about a rate. $\endgroup$ – Batman Mar 24 '15 at 21:44
  • $\begingroup$ Is the question better now, @Batman? $\endgroup$ – Helmut Mar 24 '15 at 22:10
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Since $$ \bigg| \frac1{f(n)}-\frac1a \bigg| = \frac1{|af(n)|} \cdot |f(n)-a|, $$ as soon as you know the order of magnitude of $|f(n)-a|$ and $|f(n)|$, you know the order of magnitude of $\big| \frac1{f(n)}-\frac1a \big|$.

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  • $\begingroup$ This almost makes sense to me, but your expression still has a $1/af(n)$. In the case where $a=1$, for instance, you are still left with $1/f(n)$, which was the original question. So I'm not sure this is a complete answer. Am I missing something? $\endgroup$ – Helmut Mar 24 '15 at 23:51
  • $\begingroup$ It seems actually that you've showed that the rates are the same, since assuming $a\neq 0$, $1/|af(n)|$ converges to a nonzero constant. $\endgroup$ – Helmut Mar 25 '15 at 0:12
  • $\begingroup$ Aha, good point. Yes, the rates are the same then, up to that factor $1/a^2$. $\endgroup$ – Greg Martin Mar 25 '15 at 0:31

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