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MathGroup Archive 1995

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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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