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Hello I just wanted to confirm that $$\lim_{n\to \infty} \binom{n}{3}\frac{1}{n^{3.3}} =0.$$ I have that $\binom{n}{3}$ is asymptotic to $\frac{n^3}{6}.$ So
$$ \lim_{n\rightarrow \infty} \binom{n}{3}\frac{1}{n^{3.3}} = \lim_{n\rightarrow \infty} \frac{n^3}{6}\frac{1}{n^{3.3}} = \lim_{n\rightarrow \infty} \frac{1}{6n^{.3}} = \frac{1}{6} \lim_{n\rightarrow \infty} \frac{1}{n^{.3}} = 0. $$

Thomas Andrews
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James
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    looks good to me – gt6989b Jun 23 '20 at 02:55
  • Look also at estimation https://math.stackexchange.com/questions/3722962/show-that-lim-n-rightarrow-infty-frac-binomnk2n-0/3722975#3722975 – zkutch Jun 23 '20 at 03:06
  • @zkutch, Isn't it the same estimation the OP used? $\binom{n}3\leqslant\frac{n^3}{3!}$? – PinkyWay Jun 23 '20 at 06:13
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    @Croissant. As I see, he used equivalence, which is also in link brought $\binom{n}{k} \sim C(k)n^k$, but in comments below. On other hand, of course, estimation is enough. – zkutch Jun 23 '20 at 07:17

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