Re: Sum of binomial coefficients.


The World Wide Wade wrote:
In article
<1160161130.878517.154040@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
twittoy@xxxxxxxxx wrote:

Thank you. But assume that we got some specific m. I don't want m to go
to infinity.
In that case, is your answer is still "no"?

Do you think C(n,0) + C(n,1) is approximately C(n,2) as n -> oo?

No, but it's approximately C(n,1) as n -> infinity.

I think I recall reading that C(n,0) + C(n,1) + ... + C(n,k) has no
closed form for general k. (I.e., k not a constant or n/2 or (n - a
constant)) In my enumerative combinatorics class, *I think* it was
pointed out that, if n is big compared to k,

C(n,0) + C(n,1) + ... + C(n,k)
= C(n,k) [1 + C(n,k-1) / C(n,k) + ... + C(n,0) / C(n,k)]
~ C(n,k) [1 + k/n + (k^2/n^2) + ... ]
= C(n,k) / (1 - k/n) = n C(n,k) / (n-k).

--- Christopher Heckman

The thing that is problematic for me is that if k is an order of n, say
n/4, there are too meny addents in the sum. And I wonder if still the
addends are small and thus: C(n,0)+C(n,1)+C(n,2)+...+C(n,k-1), is
approximately C(n,k). In approximately I mean up to constant
(multiplicative) factor. Or not?

Thanks again,
Yochai.


A N Niel wrote:
In article <1160139781.416149.228500@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<twittoy@xxxxxxxxx> wrote:

Hello,

I tried to figure out something regarding partial sum of binomial
coefficients with no success.

My problem is related to this form of (partial) sum:
C(n,0)+C(n,1)+C(n,2)+...+C(n,k-1),
where k is smaller than n/2.

I want to know if this sum is "theta" of C(n,k). That is, is it correct
to say that the above sum equals to C(n,k) up to constant factor?

Thanks in advance,
Yochai.


no, the sum C(2m,0)+C(2m,1)+...+C(2m,m-1) divided by C(2m,m) goes to
infinity as m goes to infinity.

.

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