> Date: Thu, 12 Feb 98 04:12:12 UT
> From: "andrew kliman" <Andrew_Kliman@CLASSIC.MSN.COM>
> To: ope-l@galaxy.csuchico.edu
> Subject: RE: [OPE-L] Addendum, Marx and historical costs
I had written:
> "I think I made a mistake here, because the term
>
> b^(-t)
> ------ *{b + b^2 + ... + b^t}
> b
>
> would not be equal to:
>
>
> {(1/b) + (1/b)^2 + ... + (1/b)^t}
Andrew replied:
> I think they *are* equal. In the second case the series in braces
> {} is written in reverse order. For instance, taking the last
> term in the first series, b^t, we get
>
> b^(-t) b^t
> ------*--- = 1/b
> b
Alejandro, now:
OK. Just for the record, the step is as follows:
(Note that I add the term b^(t-1) to the series, which makes the
procedure clearer.)
b^(-t)
{b + b^2 + ... + b^(t-1) + b^t} * -------- =
b
{1 + b + ... + b^(t-2) + b^(t-1)} * b^(-t) =
b^(-t) + b^(1-t) + ... + b^(-2) + b^(-1) =
1 1 1 1
---- + -------- + ... + ---- + ---
b^t b^(t-1) b^2 b
which is Andrew''s series.
Thank you very much!
Alejandro Ramos
P.S. I''ll try to reply the interesting Fred''s post of Feb. 13, 1998
in these days. (I''m very busy packing my papers and other
things, because we are moving to Belize City.)