Beatty Log Series - Correction

[Paul C. Leopardi]

Sorry, a correction of my own below.
On Sat, 16 Jul 2005 01:29 pm, Paul C. Leopardi wrote:
> 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)](1/n-n).
> which will give a divergent sum when t does not equal 1.