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.

You might try this: take
$$
\eqalign
{
x_1&\textstyle=( \rlap{1}\quad , \rlap{1\over2}\quad,\rlap{1\over3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr
x_2&\textstyle=( \rlap{\alpha_1}\quad , \rlap{1\over2}\quad,\rlap{1\over3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr
x_3&\textstyle=( \rlap{0}\quad , \rlap{\alpha_2}\quad,\rlap{1\over3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr
x_4&\textstyle=( \rlap{0}\quad , \rlap{0}\quad,\rlap{\alpha_3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr
}
$$
$$
\vdots
$$
for appropriately chosen scalars $\alpha_i$. Then normalize.
Note $e_1\in\text{span}\{x_1, x_2\}$, $e_2\in\text{span}\{x_1, x_2, x_3\}$,
 $e_3\in\text{span}\{x_1, x_2, x_3,x_4\}$, $\ldots\,$.
