Here's a simple algebraic solution.
The trick is to consider a factor (1-j+m)
instead of just a factor j
, and then subtract the access part (1+m)
.
Here we go
The sum being immediately evaluated is
s0[n_, m_] := Sum[(1 - j + m)^(n - 1), {j, 1, m - 1}]
s0[n, m]
(* Out[6]= -0^(-1 + n) + (-1)^(1 + n) HurwitzZeta[1 - n, -m] *)
The problem of the OP is the observation that the sum with the factor j
sjU[n_, m_] := Sum[(j) (1 - j + m)^(n - 1), {j, 1, m - 1}]
is returned unevaluated.
Now consider that the sum s0[n+1]
which has an additional factor (1 - j + m)
s0[n + 1, m] == Sum[(1 - j + m) (1 - j + m)^(n - 1), {j, 1, m - 1}]
(* Out[8]= True *)
is evaluated.
Noticing that
(1 - j + m) == -j + (1 + m)
the value sjV
of our unevaluated sum sjU
can be written as a difference
sjV[n_, m_] := -s0[n + 1, m] + (1 + m) s0[n, m]
Comparison between sjU
and sjV
And @@ Table[sjU[n, m] == sjV[n, m] // FullSimplify, {n, 1, 10}]
(* Out[10]= True *)
shows agrrement.
Discussion
It is interesting that the sum with the upper summation index extended from m-1
to m
s1[n_, m_] := Sum[(1 - j + m)^(n - 1), {j, 1, m }]
$\text{Out: }\sum _{j=1}^m (-j+m+1)^{n-1}$
is returned unevaluated (Version 10.1.0).
While shifting the upper index further, to m+1
, works
Sum[(1 + m - j)^(n - 1), {j, 1, m + 1}]
(* Out[25]= (-1)^(1 + n) (HurwitzZeta[1 - n, -m] - Zeta[1 - n]) *)
Hence only the sum ending in the seemingly harmelss term 1^(n-1)
is not evaluated.
n
to a positive integer (and not setting a value form
) does result in a symbolic result. $\endgroup$m
to a positive integer). $\endgroup$1-j+m
byk
givesSum[(m - k + 1) k^(n - 1), {k, 2, m}]
which evaluates to-m + (1 + m) HarmonicNumber[m, 1 - n] - HarmonicNumber[m, -n]
$\endgroup$