In the chapter 6.4 on normal and self-adjoint operators, there is an example of an infinite dimensional inner product space H that has a normal operator but that has no eigenvectors.

The space is the set of functions f_n(t) = e^(int) , t in [0,2pi] with inner product = 1/2pi * integral_0_2pi(e^(-int) e^(imt))dt

The operator T(f_n) = f_(n+1)

My question is what is the basis for the validity of the following statement.

any vector f in the inner product space can be represented as f = sum_i=n_to_i=m(a_i * f_i), a_m <> 0

They use this fact to prove that the operator as no eigenvectors. But it seems to imply that any member of an infinite dimensional inner product space can be represented with a finite number of basis elements, since n and m are finite.

Are there no elements of an infinite dimensional inner product space that require an infinite number of terms to represent?