orthonormal basis in $l^{2}$ I need an orthonormal basis in $l^{2}$. One possible choice would be to take as such the sequences $\{1,0,0,0,...\}, \{0,1,0,0,...\}, \{0,0,1,0,...\}$, but I need a basis where only finitely many components of the basis vectors are zero. Does anyone know a way to construct such a basis? One possible vector for such a basis would be $\{1,1/2,1/3,1/4,...\}$ devided by its norm. However, I don't know how to find similar vectors that are orthogonal to this one and to each other.

Yes, the idea of David is nice! Unfortunately, it is not what I was looking for. The problem is that the coefficients $\alpha_k$ grow to infinity when $k \rightarrow \infty$. This is the price to pay for the zeros in the beginning. Even after normalization the new coefficients, say $\widetilde{\alpha}_{k}$, will tend to $1$ and my wish was to have something that tends to zero. My initial idea was to find some permutation that does the job. For example, if I interchange the first with the second entry, the third with the fourth and so on and put a minus sign on the odd components, I get $(-\frac{1}{2},\frac{1}{1},-\frac{1}{4},\frac{1}{3},-\frac{1}{6},\frac{1}{5},...)$ which is orthogonal to $(\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},...)$ with respect to the usual scalar product in $l^2$. However, I don't know how to construct in a systematic way a third vector with these numbers that is orthogonal to the other two.
