Is this a bug?
Sum[Binomial[2 n - 1, 2 m - 1]*(L - 2 - 2*(2 m - 1)), {m, 1, n}]
gives
-4^(-1 + n) (1 - L + 2 n)
However, clearly, if n=1, the series is L-4 where m can take the only value of 1.
But the answer computed by Mathematica -4^(-1 + n) (1 - L + 2 n) with n=1 simplifies to L-3?
I am using MMA 14.0 on Windows.
You are right in pointing out that the result produced by Mathematica is only correct for n>1. This seems to comes from the term Sum[Binomial[2 n - 1, 2 m - 1](2 m - 1)), {m, 1, n}]. One can evaluate this by writing out the binomial expansions of the derivatives of (1 + x)^(2n - 1) and (1 - x)^(2n - 1), and evaluating the difference at x = 1. The issue arises from the second term which has derivative -(2n - 1)(1 - x)^(2n - 2). The Mathematica result assumes n != 1 and sets this term to 0. However, in the case n = 1, (1 - x)^(2n - 1) = 1 - x, which has derivative -1. For n != 1, say n = 2, (1 - x)^(2n - 1) = 1-3x+3x^2-x^3, the derivative is -3 + 6x - 3x^2 for which the value is 0 at x = 1 and this result holds for all n > 1.
If you enter Sum[Binomial[2 n - 1, 2 m - 1]*(L - 2 - 2*(2 m - 1)), {m, 1, n}, Assumptions -> n == 1], Mathematica does produce L - 4, but not for assumptions such as n >= 1, where the answer you provided is the result.
$Version
(* "14.0.0 for Mac OS X ARM (64-bit) (December 13, 2023)" *)
Clear["Global`*"]
sum1[n_Integer?Positive] :=
Sum[Binomial[2 n - 1, 2 m - 1]*(L - 2 - 2*(2 m - 1)), {m, 1, n}]
sum1[1]
(* -4 + L *)
Generating a sequence,
seq = sum1 /@ Range[8] // Simplify
(* {-4 + L, 4 (-5 + L), 16 (-7 + L), 64 (-9 + L), 256 (-11 + L), 1024 (-13 + L),
4096 (-15 + L), 16384 (-17 + L)} *)
Using FindSequenceFunction
sum2[n_] = FindSequenceFunction[seq, n]
The result is a DifferenceRoot
sum2[n] // InputForm
(* DifferenceRoot[Function[{\[FormalY], \[FormalN]},
{16*\[FormalY][\[FormalN]] - 8*\[FormalY][1 + \[FormalN]] +
\[FormalY][2 + \[FormalN]] == 0, \[FormalY][1] == -4 + L,
\[FormalY][2] == 4*(-5 + L), \[FormalY][3] ==
16*(-7 + L), \[FormalY][4] == 64*(-9 + L),
\[FormalY][5] == 256*(-11 + L),
\[FormalY][6] == 1024*(-13 + L),
\[FormalY][7] == 4096*(-15 + L)}]][n] *)
Verifying that the DifferenceRoot works beyond the original sequence:
And @@ (Table[sum1[n] == sum2[n], {n, 1, 50}] // Simplify)
(* True *)
GenerateConditions -> Truegives the same thing along withn >= 1which is a bug, but when you dont add that option i wouldn't say a return that works forn > 1is a bug. $\endgroup$SumandIntegrateoften give results that whose dependence on parameters is only "generically true" (finite number of exceptions when the parameter is a positive integer). Might always want to check the first couple of values. $\endgroup$