Determine if $g=\sum_{n=1}^\infty n^{-1/3}e_n$ exists in Hilbert space

Question

In a Hilbert space I have an orthogonal basis $(e_n)_{n=1}^\infty$

I want to find out if an element $g \in \mathcal{H}$ is defined using:

$g=\sum_{n=1}^\infty n^{-1/3}e_n$

I think this is only possible if the sum converges. However I don't know how to handle the converge. For a p-series it would not converge because we require $p>1$. But here we are also multiplying with $e_n$ so I am not sure how this affects the sum because each term is a separate coordinate in g

Any hint would be appreciated

Answer 1

Given a sequence of real numbers $\left(a_n\right)_{n\geqslant 1}$, can we define $$ v=\sum_{n=1}^{+\infty}a_ne_n $$ by $v=\lim_{N\to +\infty}v_N$ where $v_N:=\sum_{n=1}^{N}a_ne_n$?

Since a Hilbert space is complete, this is doable if and only if the sequence $\left(v_N\right)_{N\geqslant 1}$ is Cauchy for the norm induced by the inner product. Since for fixed $M$ and $N$, $\lVert v_{N+M}-v_N\rVert^2=\sum_{n=N+1}^{N+M}a_n^2$, the necessary and sufficient condition is the convergence of $\sum_{n=1}^{+\infty}a_n^2$.

This is in particular not satisfied for $a_n=n^{-1/3}$.

Answer 2

Hint: $\displaystyle\left\lVert\sum_{n=1}^Nn^{-1/3}e_n\right\rVert^2=\sum_{n=1}^Nn^{-2/3}$.