a binomial coefficient approximation is needed.

Question

sumeer · Answer

here is this.

kainui · Answer

What do you need your approximation for? This seems pretty good.

You can approximate factorials with Sterling's approximation, and plug those in to the binomial I guess too.

sumeer · Answer

I should say I need to show this approximation. I attach what I did.

kainui · Answer

Oh, you probably shouldn't use Sterling's approximation if you want to prove that inequality.

sumeer · Answer

I still havent found the solution..  :/

divinesolar · Answer

Multiplicative formula? https://gyazo.com/9fbb96f73581654eaef0c52143695239

anthony1914 · Answer

still need help

dannyo19 · Answer

Remind me what the symbol before the n means?

sumeer · Answer

@Danny019 are you mentioning $$(1- \frac{ k}{ 2n })$$

reemii · Answer

By induction. For any $n\ge 0$, we have: $\forall k=0,1,\dots,n$, $\binom{n}{k} \le \dotsb$

Fix, $n=N+1$, then
1) $\binom{N+1}{0} \le 1$ : ok.
2) the other cases: we have $\binom{n+1}{k+1} = \frac{n+1}{k+1}\binom{n}{k}$. This and $(1-k/(2n)) = (2n - k)/(2n)$ should work.

reemii · Answer

* other cases: we have $\binom{N+1}k+1{}$ etc.
-> it's $n$ or $N+1$, not $n+1$.

reemii · Answer

* $\binom{N+1}{k+1}$ ...

sumeer · Answer

Its a clever approach.There are 3 cases to consider, one is n gets one bigger and k is fixed, and second is vice versa, and third is both gets one bigger. Proving for these cases finishes the proof. Thanks from my heart @reemii

reemii · Answer

I see it just as a 2-step procedure at most. It's a conventional induction proof. It's just that the step $n=N+1$ requires to verify $k+1$  inequalities (this is done by considering two cases).

reemii · Answer

Congratulations for finishing the proof

sumeer · Answer

Thanks.