Re: Sums
- To: mathgroup at smc.vnet.net
- Subject: [mg28044] Re: [mg27996] Sums
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Thu, 29 Mar 2001 03:24:10 -0500 (EST)
- References: <200103280740.CAA25681@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Janusz Kawczak wrote:
>
> 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.
To handle that IntegerPart you might rewrite as a doubly indexed sum, as
below.
In[25]:= InputForm[Sum[Sum[(j+k*n)*(1-r)^n*Boole[0<r<.5],
{n,0,Infinity}], {j,0,k-1}]]
Out[25]//InputForm= -(k*(-2*k + r + k*r)*Boole[0 < r < 0.5])/(2*r^2)
Daniel Lichtblau
Wolfram Research
- References:
- Sums
- From: Janusz Kawczak <jkawczak@math.uncc.edu>
- Sums