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Re: Sums

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28059] Re: [mg27996] Sums
  • From: Tomas Garza <tgarza01 at prodigy.net.mx>
  • Date: Thu, 29 Mar 2001 03:24:24 -0500 (EST)
  • References: <200103280740.CAA25681@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Ok, I propose the following for the first problem. I assume k is integer >1.
I take the first few terms of  each of the series for k = 2, 3, 4, 5, and
take the finite sum (say the first 40 terms). Observe the coefficients (I
use q = 1 - r to make iit a less cumbersome).

In[1]:=
Table[CoefficientList[Plus@@ Table[n*(1 - r)^IntegerPart[n/k], {n, 1, 40}]
/.
   (1 - r -> q),q],{k,2,5}]
Out[1]=
{{1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73,
77,
     40}, {3, 12, 21, 30, 39, 48, 57, 66, 75, 84, 93, 102, 111, 79}, {6, 22,
    38, 54, 70, 86, 102, 118, 134, 150, 40}, {10, 35, 60, 85, 110, 135, 160,
    185, 40}}

It is evident that in all cases they follow an arithmetic progression (which
should be clear due to the effect of IntegerPart[n/k]), except for the last
term, and this is so because of the truncation effect. The first term is
always equal to Binomial[k,2], and we find, in general, that the constant
difference is k^2, so that the n-th term of the series for each k is given
by

In[2]:=
<<DiscreteMath`RSolve`

In[3]:=
c[k_] := RSolve[{a[n] == k^2 + a[n - 1], a[1] == Binomial[k, 2]}, a[n], n]

Then, for example, k = 5 gives

In[4]:=
c[5]
Out[4]=
{{a[n] -> 5 If[n >= 1, -3 + 5 n, 0]}}

so that the corresponding sum of the series for k = 5 is

In[5]:=
Binomial[k, 2] + Sum[-5*(-3 + 5*n)*q^n, {n, 1, Infinity}]/.q->(1-r)
Out[5]=
6 + (5*(2*(1 - r) + 3*(1 - r)^2))/r^2

I guess the whole thing could be automatized in a single shot. The finite
sums can be obtained with a little additional labor.

Tomas Garza
Mexico City


----- Original Message -----
From: "Janusz Kawczak" <jkawczak at math.uncc.edu>
To: mathgroup at smc.vnet.net
Subject: [mg28059] [mg27996] Sums


> Hello:
>
> can anybody help? I am trying to evaluate the following:
>
> Sum[n (1-r)^(IntegerPart[n/k]) Boole[ 0 < r < .5], {n, 1, Infinity}],
> for some k >1, but fixed.
>
> In the same fashion, many other sums are of interest:
>
> Sum[n (n+1) (1-r)^(IntegerPart[n/k]) Boole[ 0 < r < .5], {n, 1,
> Infinity}],
> for some k >1, but fixed, and so on.
>
> Also, the finite sums are of some interest to me as well, i.e.
>
> Sum[n (1-r)^(IntegerPart[n/k]) Boole[ 0 < r < .5], {n, 1, N}],
> for some k >1, but fixed.
>
> The above sum can be easily evaluated without the IntegerPart in
> the exponent. But, I cannot get it to work in general.
>
> Any help with this matter will be greatly appreciated.
> John.
>
> P.S. Please cc to jkawczak at math.uncc.edu.
>
>



  • References:
    • Sums
      • From: Janusz Kawczak <jkawczak@math.uncc.edu>
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