Abel's Convergence Theorem exercise

I am trying to answer the exercise in my book, but I get an incorrect
result. (I'm probably carrying over some asumptions from real analysis
that I need to shake out). Can any one suggest where I went wrong?

Question:
Given that SUM(0,inf)a_k is convergent, a_k in C. Let r>=1, and let
E={|1-z|<=r(1-|z|)}. We have E subset Delta(0;1)U{1}.
Show that f is continuous at 1, f being SUM(0,inf)a_kz^k on E.

Hint:
f(1) - f(z) = SUM(1,inf)a_k(1-z^k) = SUM(1,m)a_k(1-z^k) +
SUM(m+1,inf)a_k(1-z^k)
Apply the Abel identity to SUM(m+1,m+p)a_k(1-z^k) with the (1-z^k)
being the a's of the identity.

(Attempted) Answer:
ok, so I started with:
SUM(m+1,m+p)a_k(1-z^k)

Applying Abel's identity (this was a little tricky), I got:

SUM(m+1,m+p)a_k(1-z^k) = SUM(m+1,m+p-1){(1-z^k) - (1-z^(k+1))}S_k +
(1-z^(m+p))S_n

where S_k = SUM(m+1,k)a_k.

Simplifying, I get:
SUM(m+1,m+p-1){z^(k+1)-z^k}S_k + (1-z^(m+p))S_n

= (z-1)SUM(m+1,m+p-1)z^ks_k + (1-z^(m+p))S_n

Wasn't exactly sure what to do next, so I took the absolute value:

|(z-1)SUM(m+1,m+p-1){z^kS_k} + (1-z^(m+p))S_n|

I note that |s_k|<ep (epsilon) for big enough m, and moved it to the
outside (is that allowed?):

< ep.|(1-z)SUM(m+1,m+p-1)z^k + (1-z^(m+p))| =
ep.|(1-z).z^(m+1).(1-z^p)/(1-z) + (1-z^(m+p))|

= ep.|z^(m+1).(1-z^p) + (1-z^(m+p))| <= 4ep for |z|<=1.

But this would yield a set E={|z|<=1, z!=1}

Which doesn't seem correct? Help!

.