I guess this is a bug.
But, in case, you want a quick fix, you might want to write a program by
working this sum. This is not hard, once one realises that, one could
rearrange the definition of the Beta function and observe ($n geq k$):
$$\binom{n}{k}^{-1} = \int_0^1 t^k (1-t)^{n-k} dt$$
This should prove that the sum you're after is:
n+1/2^{n+1} sum(2^k/k, k, 1, n+1).
Hope this helps.
PS. If somebody can confirm that it is really a bug (not really familiar
with symbolic...), then, I'd open a ticket on trac.
On Thu, Jul 18, 2013 at 7:17 AM, Ed Scheinerman
edward.scheiner...@gmail.com wrote:
I am looking to evaluate $\sum_{k=0}^n 1/\binom{n}{k}$ so I type:
sage: n = var('n')
sage: k = var('k')
sage: sum(1/binomial(n,k),k,0,n)
(n + 1)*2^(-n)
and that answer is wrong.
For example, with n=10 we get
sage: sum(1/binomial(10,k) for k in range(11))
1433/630
but the alleged answer of (n+1)*2^(-n) is 11/1024.
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