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Re: Help on Infinite Series
*To*: mathgroup at smc.vnet.net
*Subject*: [mg2451] Re: Help on Infinite Series
*From*: Bob Hanlon <hanlon at pafosu2.hq.af.mil>
*Date*: Wed, 8 Nov 1995 23:45:37 -0500
Using Mathematica:
Needs["Algebra`SymbolicSum`"]
PROBLEM: Find Sum[k^m, {k, 1, n}]
f1[n_, m_] := Evaluate[SymbolicSum[k^m, {k, 1, n}]];
f1[n, m]
Zeta[-m] - Zeta[-m, 1 + n]
Table[f1[n, m], {m, 0, 4}] // TableForm
n
1
-(--) - Zeta[-1, 1 + n]
12
-Zeta[-2, 1 + n]
1
--- - Zeta[-3, 1 + n]
120
-Zeta[-4, 1 + n]
Except for the trivial case (m = 0), the results are not readily
understandable since they are left in the form of generalized Zeta
functions. However, an alternate -- indirect -- approach provides the
results in a more useful form. Generalizing the sum to a power series in
x:
g1[n_, m_, x_] :=
Evaluate[SymbolicSum[k^m * x^k, {k, 1, n}]]
g1[n, m, x]
n
x (LerchPhi[x, -m, 1] - x LerchPhi[x, -m, 1 + n])
g1[n, m, 1]
Zeta[-m] - Zeta[-m, 1 + n, IncludeSingularTerm -> False]
This is the same result; however, since,
x * D[g1[n, m, x], x] = Sum[k^(m+1) * x^k, {k, 1, n}] = g1[n, m+1, x]
then, g1 can be defined recursively. The starting point for the recursion
is
g1[n, 0, x] // Simplify
n
x (-1 + x )
-----------
-1 + x
g2[n_, 0, x_] := x * (1 - x^n) / (1 - x);
g2[n_, m_, x_] := g2[n, m, x] =
Simplify[x * D[g2[n, m-1, x], x]];
mmax = 7;
Table[g2[n, m, x], {m, 0, mmax}]; (* long output suppressed *)
f2[n_, m_] := f2[n, m] = Limit[g2[n, m, x], x->1];
Table[f2[n, m] // Factor, {m, 0, mmax}] // TableForm
n
n (1 + n)
---------
2
n (1 + n) (1 + 2 n)
-------------------
6
2 2
n (1 + n)
-----------
4
2
(-1 - n) n (1 + 2 n) (1 - 3 n - 3 n )
-------------------------------------
30
2 2 2
n (1 + n) (-1 + 2 n + 2 n )
-----------------------------
12
3 4
n (1 + n) (1 + 2 n) (1 - 3 n + 6 n + 3 n )
-------------------------------------------
42
2 2 2 3 4
n (1 + n) (2 - 4 n - n + 6 n + 3 n )
----------------------------------------
24
This agrees with published results (e.g., Hansen, Eldon R.; A Table of
Series and Products, Prentice-Hall, New Jersey, 1975, Section 5.14)
Checking:
And @@ Flatten[
Table[f1[n, m] == f2[n, m], {n, 0, 7}, {m, 0, 7}]]
True
> From: mfischer at trincoll.edu
To: mathgroup at smc.vnet.net
> Newsgroups: sci.math.symbolic
> Subject: Help on Infinite Series
> Date: Mon, 6 Nov 1995 19:38:41
>
> I know that the summation of n squared can be described as
>
> n(n+1)(2n+1) / 6
>
> does anyone know how to prove this?
>
> I know that the difference in the differences in the series is
>
> 2n+1 but I don't know how to get the rest
>
> Also how would you describe n-cubed?
>
> the difference of the difference of the differences is
>
> 6(n+1) but I don't know how to encorporate that
>
> any help would be greatly appreciated
>
> please mail any help to me at fischer at micro5.cs.trincoll.edu
>
> thanks in advance
>
> matt
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