Show that $\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0$

Question

Show that the following limit holds $$ \lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0 $$ for a fixed value of $k$

I really am just stuck at the first step here. Normally I would consider tackling this using L'Hopitals rule, however $\binom{n}{k}$ is not differentiable. I was considering using the binomial theorem, but that is for sums, not just the single scenario. Any help appreciated!

I didn't try it but I would suggest writing $n \choose k$ $= \frac{n!}{k! (n-k)!}$ and use Stirling's formula to estimate the factorial. — mathdoge, Jun 17 '20 at 04:28
Does this answer your question? Prove or disprove that $\lim_{n \to \infty} {n \choose k} / a^n = 0$ for all $k \in \mathbb{Z_+}, a \in \mathbb{R}: a> 1$. — Martin R, Jun 17 '20 at 04:46
Its not as rigourous as I'd like, but I think its a good enough way for me to think about it, and provides a bound so that I can do out the computations. Thank you — wjmccann, Jun 17 '20 at 04:51
@wjmccann Note your question, as I state in my comment to zkutch's answer, involves basically showing there are quite a few posts here that explain, in various ways, that exponentials grow faster than polynomials, e.g., Is 'every exponential grows faster than every polynomial?' always true? and Proving exponential is growing faster than polynomial. — John Omielan, Jun 17 '20 at 04:58

zkutch · Accepted Answer · 2020-06-23T09:10:21.863

3

If you accept as known $\binom{n}{k} \leqslant \frac{n^k}{k!}$, then you obtain it by estimation.

Addition: added second part, as it can be helpful for somebody. $$\frac{n^k}{k^k} \leqslant \binom{n}{k} \leqslant \frac{n^k}{k!}$$

edited Jun 23 '20 at 09:10

answered Jun 17 '20 at 04:41

zkutch

13,410

1

But even if you don’t accept this, it suffices to accept that it is less than $n^k$. This is indeed true because $n^k$ is the number of ways to pick $k$ objects from $n$ objects with repetition allowed and the order matters. This certainly overcounts the number of ways where we can pick $k$ objects from $n$, but where there are no repetitions and the order doesn’t matter (which is what ${n \choose k}$ represents). – paulinho Jun 17 '20 at 04:45
@zkutch Note this involves the basic case where exponentials (i.e., $2^n$) grow faster than any polynomial (i.e., $\frac{n^k}{k!}$), which you can also prove using L'Hôpital's rule on the numerator & denominator $k$ times. – John Omielan Jun 17 '20 at 04:46
@John Omielan. Sorry, I miss your idea - of course exponent grow faster than any polynomial, but what you want to say? I put simplest way I know. – zkutch Jun 17 '20 at 04:57
@zkutch I was just pointing out not only for yourself, but also for anybody else reading it, a few extra possibly useful details. I didn't mean to imply anything was wrong with your answer, and I also wasn't asking or expecting you to say anything in particular, which is why I started my comment with "Note". – John Omielan Jun 17 '20 at 05:02
@John Omielan. Thanks for explanation. Let me, also from my side, add for such people, that it is possible to say more: $\binom{n}{k} \sim C(k)n^k$ where $C(k)$ depends only on $k$. – zkutch Jun 17 '20 at 05:06

score -2 · Answer 2 · answered Jun 17 '20 at 04:36

-2

Set k = n since it is the greatest it will ever be. The numerator will never surpass 1. So you simply have the ratio of 1 to infinity.

answered Jun 17 '20 at 04:36

George Ntoulos

188

By $\binom{n}{k}$ I mean "$n$ choose $k$", not $\frac{n}{k}$. Sorry for any confusion! – wjmccann Jun 17 '20 at 04:39
$k = n$ does not attain the maximum... – mathdoge Jun 17 '20 at 04:40
Max is at $k=n/2$ – Integrand Jun 17 '20 at 04:41
@mathdoge I misunderstood the question. Didn't read carefully and I was half asleep. – George Ntoulos Jun 17 '20 at 07:22

Show that $\lim_{n\rightarrow\infty} \frac{\binom{n}{k}}{2^n} =0$

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