Determining if a set is a Basis for l^2 For each $ n\ge 1$ Define the vectors $e_n = (e_{nk})$ where  $ k\ge 1$ and  $ e_{nk} = \frac{1}{k^n}$
Is this set a basis for $l^2$?
Thanks,
 A: You want to know if $\sum_{k=1}^\infty a_k k^{-n}=0$ for $(a_k)\in l^2$ and every positive integer $n$ implies $a_k=0$. This is true. First we note that if $(a_k)\in l^2$, then $(a_k/k)\in l^1$. So w.l.o.g. we may consider the case where $(a_k)\in l^1$ to begin with. Now note that the condition implies that $\sum_{k=1}^\infty a_kp(1/k)=0$ for every polynomial $p$ with $p(0)=0$. Now use the Weierstrass approximation theorem to conclude that $\sum_{k=1}^\infty a_k f(1/k)=0$ for every continuous function $f$ on $[0,1]$ with $f$ vanishing near the origin. It follows easily that $a_k=0$.
A: A slightly more general approach which uses less of the structure on the coefficients, showing that the span of the $(e_n)$'s is dense in $\ell^2$. If $a$ is orthogonal to all of the $e_n$'s then
$$ a_1 = \lim_n (a,e_n) = 0$$
and then $$ a_2 = \lim_n \sum_{k\geq 2}  \frac{2^n}{k^n} a_k=\lim_n  2^n (a,e_n) =  0$$
etc...
A: This is like a Müntz-Százs Theorem for series. See this survey http://arxiv.org/abs/0710.3570
Here the usual tool of Gram determinants gives values of the zeta function as matrix entries which does not seem too promising. I would try to project on the subspace corresponding to $k$
a power of a fixed prime $p$. If the projections are not total in the subspace, neither are your $e_n$'s.
Answering this type of questions about a specific set being total in $l^2$
can be extremely hard, see: http://link.springer.com/article/10.1007/BF02829783
