sum involving binomial coefficient

Question

Bug introduced in 9.0 or earlier and persisting through 11.0.1 or later

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To evaluate $ \sum_{\substack{\text{even }k\\ 2\leq k\leq m-2}} \binom{m-2}{k-1}(k-1) $, where $m$ is an even integer.

Running the following in Mathematica 10.3.0 (October 9, 2015)

Simplify[Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m-2, 2}], m/2 \[Element] Integers && m > 10]

gives

0

which is obviously wrong. It may not have a closed form but it shouldn't return 0.

Answer 1

This is a bug in Sum.

In addition to the nice workarounds proposed by others, the substitution $m = 2p$ also leads to a correct answer for this example.

In[1]:= Sum[
   Binomial[2 p - 2, k - 1] (k - 1), {k, 2, 2 p - 2, 2}] /. {p -> 
    m/2} // Simplify


Out[1]= 2^(-4 + m) (-2 + m)

Sorry for the inconvenience caused by this problem.

Answer 2

I would suggest defining your sum as a function

f[m_?NumericQ] := Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}] /; EvenQ[m];

This evaluates fine and does not give zero when given numeric values.

The root of the problem seems to me to have to do with the definition of Binomial internaly as a combination of Gamma functions.

"In general,Binomial[n,m] is defined by [CapitalGamma] (n+1)/([CapitalGamma](m+1)[CapitalGamma](n-m+1)) or suitable limits of this."

As evidence, the expression

Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}]

Evaluates to zero.

Answer 3

As a workaround for this incorrect simplification, you could use FindSequenceFunction on a sample of concrete results:

Clear[m,n,k];
Simplify[FindSequenceFunction[
   Table[
    Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}], {m, 2, 12, 2}],
    n] /. n -> m/2]

(* ==> 2^(-4 + m) (-2 + m) *)

This is similar to the answer by @Coolwater here.