Beatty Log Series - Correction

[Paul C. Leopardi]

Paul, reply below. Best regards, Paul Leopardi
On Fri, 15 Jul 2005 03:42 pm, Paul D. Hanna wrote:
> 1/[n(1+t)] - 1/[n(1+1/t)] - (1-t)/(1+t)/n

Lets call this term b(t,n).

By (1-t)/(1+t)/n do you mean:

1. [(1-t)/(1+t)]/n
(This is the interpretation used by Maple and most programming languages)
OR
2. (1-t)/[(1+t)/n] ?

With interpretation 1, we have
[(1-t)/(1+t)]/n = (1-t)/[n(1+t)]
and thus the whole expression b(t,n) is
b(t,n) = t/[n(1+t)] - 1/[n(1+1/t)]
       = t/[n(1+t)] - t/[n(t+1)]
       = 0 assuming that t is non-zero.

With interpretation 2, we have
b(t,n) = (1-t)/[n(1+t)] - (1-t)/[(1+t)/n]
       = [(1-t)/(1+t)](n-1/n)
       which will give a divergent sum when t does not equal 1.