Showing a certain subspace of Hilbert space is dense Let H be the Hilbert space of square-summable sequences of reals.
A few years ago I thought I had proved that the subspace Z of real sequences with only finitely many nonzero terms, such that they sum to zero, is dense in H. 
(Since then I've seen this confirmed in Rudin's text, Functional Analysis, 2nd ed., but only as a teensy subquestion in a terminally hairy exercise.)
Now I can't quite reproduce my proof, so it may have been wrong. Can someone please tell me how to prove that.  
(What I tried was taking an arbitrary point c = (c1,...,ck,...) of H, and defining the point d(N) in H as follows for N $\ge$ 1:
First set AN = (c1+...+cN) / N. Then set d(N)k =  ck - AN for k $\le$ N, and set d(N)k = 0 for k > N.
Clearly, the element d(N) of H lies in the subspace Z defined above. The squared Hilbert norm of its difference with c is of form N(AN)2 + T(N), where T(N) is just the squared norm of the tail of c, and so goes to 0 as N$\to\infty$.
I'm left with the expression N(AN)2, which so far I haven't been able to prove $\to$ 0 as N$\to\infty$.  I suspect this is true, and it works on all the examples I've tried so far.)
NOTE: I'm only interested in a down-and-dirty proof that doesn't invoke anything but simple inequalities.
Thanks for any help you can offer.

Given $x \in H$ and $\epsilon > 0$, there is $y \in H$ with $\|x - y\| < \epsilon/2$ such that only finitely many elements of $y$ are nonzero.  Suppose the sum of those nonzero elements of $y$ is $s$.  Consider $z$ obtained from 
$y$ by changing $n$ of its elements from $0$ to $-s/n$.  Thus $z \in Z$,
and $\|z - y\|^2 = n (s/n)^2 = s^2/n$.  If $n$ is sufficiently large, this is
less than $(\epsilon/2)^2$.  So $\|x - z\| < \epsilon$.  We conclude that $Z$ is dense in $H$.
