0

$C^{m}_{n}=\frac{n!}{m!(n-m)!}$ are the binomial coefficients. Now for a fix $0<\alpha<1$, consider for any natural number $N$ the following sequence: $$ T_N=\sum_{[\alpha N]\leq i\leq N}C^i_{N}$$ Could you help me with estimating the limit of $T_N/2^N$ when $N\to \infty$

metamorphy
  • 43,591
Tibeku
  • 374
  • I recall this basically-the-same question (with my answer there giving a little bit finer result for $\alpha\neq1/2$). But the appearance of the sum makes me suspect that this is asked many times. – metamorphy Jan 16 '21 at 09:48

1 Answers1

1

$$T(n)=\sum _{i=\lfloor n \alpha \rfloor }^n \binom{n}{i}=\frac{\Gamma (n+1) \, _2F_1(1,n (\alpha -1);n \alpha +1;-1)}{\Gamma (n-\alpha n+1) \Gamma (n \alpha +1)}$$

From numerical results I got that $$\underset{n\to \infty }{\text{lim}}\frac{T(n)}{2^n}=\begin{cases} 1; & 0<\alpha<\frac12\\ \frac12;& \alpha=\frac12\\ 0; &\frac12<\alpha<1\\ \end{cases}$$

Raffaele
  • 26,731
  • I don't understand: according to your formula, $T(n) \leq 1$, which is apparently untrue. – Tibeku Jan 11 '21 at 22:18
  • I misread you sum and inverted $i$ and $n$ because in $C^i_{N}$ I read $i$ high and $N$ low, like in $\binom{i}{N}$. – Raffaele Jan 11 '21 at 22:39