D=26 in bosonic string theory?

[Calvin D. Ritchie]

I'm just getting started on reading Polchinski, "String Theory", V 1,
and have run into something that leaves me totally lost. I've looked
at Polchinski's web site to see if there is, perhaps, a correction,
but found nothing pertinent. I have the 2005 printing of the book
which seems to include most of the corrections on Polchinski's
web-site.

On pg. 22, Eq. (1.3.34), Polchinski writes, essentially:

Sum(n=1, infty) n*Exp(-eps*n)=(1/eps^2) - (1/12) + Order(eps).

I've used eps= Polchinski's (epsilon*(Pi/(2p^+) *alpha'*l)^1/2, but
don't see how that can change anything essential. The r.h.s. of that
equation dumbfounds me. The sum on the l.h.s. can be written, exactly,
as

Sum(n=1, infty) n*Exp(eps*n)=[Exp(-eps)]/[1-Exp(-eps)]^2.

I don't see any approximation that leads to Polchinski's result,
particularly the 1/12 term, which is essential to his conclusion of
D=26. Moreover, if I approximate the sum by an integral, I get

Integral(n=1, infty){dn*n*Exp(-eps*n)}={(1/eps^2) +
(1/eps)}*Exp(-eps),

and I still can't see how that could possibly give Polchinski's
result. Note that a change in the lower limit of the sum or integral
from n=1 to n=0 doesn't change the sums, and only changes the integral
evaluation to -> (1/eps^2) as the only term.

Can anyone tell me what I'm missing or mis-interpreting?

Don Ritchie

[Calvin D. Ritchie]

Within an hour of sending off the question, I realized the answer.
Sorry to be so dense, but, just in case anyone else has trouble seeing
the correct approach:

Exp(-eps)/[1-Exp(-eps)]^2 =
Exp(-eps)/[1-2Exp(-eps)+Exp(-2eps)]=1/[Exp(+eps)-2+Eps(-eps)],

Now, taking the usual expansion of the exponentials in the
denominator, only even powers of eps survive, and

1/[Exp(+eps)-2+Eps(-eps)] ~ 1/[eps^2 + (eps^4)/12 + ......] ~
[1-(eps^2)/12]/{eps^2[1-(eps^4)/144 + ....] ~ 1/eps^2 - 1/12 + .......

Also, if anyone is interested, there's a better, IMO, presentation of
this development in Zwiebach's "A First Course in String Theory", pgs.
221 and Problem 12.4 on pg. 243.

Don Ritchie