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Could someone please explain the second equality in Conjecture 1.1: ? (reproduced below)


Initially, I thought that using the binomial formula on the RHS would do it - but it befuddles me how one can pull out an $n^2$ term (from $\binom{n}{2}$) and ignore the rest. Thanks!

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up vote 1 down vote accepted

For any nonnegative function $f(n)$ and any $n \geq 1$ one has \begin{align} (1 + f(n)) n^2 2^{1-n} & = \frac{2n^2 + 2 n^2 f(n)}{2^n} \\ & = \left( \frac{(2n^2 + 2n^2 f(n))^{1/n}}{2}\right)^n. \end{align} The assertion that $(1 + o(1)) n^2 2^{1-n} = (\frac{1}{2} + o(1))^n$ is thus the assertion that if $f(n) \to 0$ as $n \to \infty$, then $\frac{(2n^2 + 2n^2 f(n))^{1/n}}{2} - \frac{1}{2} \to 0$ as $n \to \infty$ (ie, that $\frac{(2n^2 + 2n^2 f(n))^{1/n}}{2} - \frac{1}{2} = o(1)$, so that $\frac{(2n^2 + 2n^2 f(n))^{1/n}}{2} = \frac{1}{2} + o(1)$).

But if $f(n)$ is any bounded function, with, say, $f(n) \leq B$ for all $n$, clearly $$ (2n^2)^{1/n} \leq (2n^2 + 2n^2 f(n))^{1/n} \leq ((2 + 2B) n^2)^{1/n}, \qquad n \geq 1, $$ and short calculations (involving, perhaps, L'Hopital's rule) show that $(2n^2)^{1/n}$ and $((2 + 2B)n^2)^{1/n}$ both go to $1$ as $n \to \infty$. We conclude that $(2n^2 + 2n^2 f(n))^{1/n} \to 1$ as $n \to \infty$, and hence $$ \frac{(2n^2 + 2n^2 f(n))^{1/n}}{2} - \frac{1}{2} \to 0, \qquad n \to \infty, $$ as desired.

(If I haven't missed something, this proves the slightly stronger $(1 + O(1)) n^2 2^{1-n} = (\frac{1}{2} + o(1))^n$, and the same argument could be adapted to prove something more general still.)

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