Combinatoric formula summing one

Question

While working in a problem derived from a mathematical model of a game, I found the following formula: $$ lim_{n\rightarrow\infty}\sum_{i=1}^m(-1)^{i-1} \binom{n-i}{i} c^{-2i}=1, $$ where $c$ is any positive integer major or equal than 2, $m=n/2$ if $n$ is even and $m=\frac{n-1}{2}$ if $n$ is odd.

I know it is correct because I have tested it in the computer, but I have no idea of how to prove it.

Answer 1

At this answer the following formula has been posted:

$$a^n+b^n=(a+b)^n-\sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{n-2i}(ab)^i$$

Factoring out $(a+b)^{n}$

$$a^n+b^n=(a+b)^n-(a+b)^{n}\cdot \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}(ab)^i$$

$$(a+b)^n-a^n+b^n=(a+b)^{n}\cdot \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}(ab)^i$$

Dividing the equation by $(a+b)^{n}$

$$1-\frac{a^n+b^n}{(a+b)^{n}}= \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}(ab)^i$$

We can set $a\cdot b=1$. Consequently $(ab)^i=1$ disappears.

$$1-\frac{a^n+b^n}{(a+b)^{n}}= \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}(a+b)^{-2i}$$

Then we have two equations:

1. $a\cdot b=1\Rightarrow a=\frac{1}b$
2. $a+b=c$

The resulting quadratic equation $\frac1b+b=c$ has a solution for $b$ only if the discriminant is non-negative. This is when $c^2\geq 4$. If confirms the condition of Claude Leibovici. Now we can look at the limit.

$$1-\lim_{n \to \infty}\frac{a^n+b^n}{(a+b)^{n}}= \lim_{n \to \infty} \sum _{i=1}^{n-1}(-1)^{i-1}\frac{n}{n-i}\binom{n-i}{i}c^{-2i}$$

$\lim\limits_{n \to \infty}\frac{n}{n-i}=1$

$$1-\lim_{n \to \infty}\frac{a^n+b^n}{(a+b)^{n}}= \lim_{n \to \infty} \sum _{i=1}^{n-1}(-1)^{i-1}\binom{n-i}{i}c^{-2i}$$

$$1= \lim_{n \to \infty} \sum _{i=1}^{n-1}(-1)^{i-1}\binom{n-i}{i}c^{-2i}$$

If we calculate at the limit ($\to$ infinity) we don´t have to distinguish between the odd- and the even-cases.

Answer 2

You are facing the problem of the limits of the gaussian hypergeometric functions.

If $n=2p$, the result of the summation is $$1-\, _2F_1\left(\frac{1}{2}-p,-p;-2 p;\frac{4}{c^2}\right)$$

If $n=2p+1$, the result of the summation is $$1-\, _2F_1\left(-p-\frac{1}{2},-p;-2 p-1;\frac{4}{c^2}\right)$$

But you need $c^2>4$