Suppose we know that

$\lim_{n \rightarrow \infty} (f_n / g_n) = 1$,

for some sequences $\{f_n\}$ and $\{g_n\}$. Does this imply

$\lim_{n \rightarrow \infty} f_n - g_n = 0$ ?

I assume that the answer is false but I have no intuition for how the first condition doesn't imply the second. Perhaps there's a good counterexample?

Context: I ask because this issue comes up when Efron (1975) explains the difference between estimators with first and second order efficiency, see Equation 1.1. Apparently $\lim_{n \rightarrow \infty} (f_n / g_n) = 1$ is a weaker condition.

Efron, Bradley. "Defining the curvature of a statistical problem (with applications to second order efficiency)." The Annals of Statistics (1975): 1189-1242.

  • 2
    $\begingroup$ This is false, just take $f_n = n$ and $g_n = n+1$. $\endgroup$ – Schmidt May 20 '18 at 3:11
  • $\begingroup$ As soon as I finished writing this I thought of $f_n = 2^n$ and $g_n = 2^{n - 1/n}$. $\endgroup$ – Anthony Ebert May 20 '18 at 3:18

Take for instance $f_n=2^n+ (-1)^n$ and $g_n=2^n$.

Then $f_n-g_n=(-1)^n$ so it is not even convergent.

One can play around, and have something like $f_n= 2^n+z_n$ where $z_n=\text{o}(2^n)$, and $g_n=2^n$.

Of course if $f_n$ convergences to a real number then $\lim_n f_n-g_n=0$


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