Let $(e_n)_{n\in \mathbb Z}$ be an orthonormal basis in $H$ indexed in $\mathbb Z$. We consider the sequence $u_n=e_{-n}+ne_n$,
$M =\overline{\operatorname{span}\left\lbrace u_n: n ≥ 1\right\rbrace}\,$ and
$\,F =\overline{\operatorname{span}\left\lbrace e_n: n ≥ 0 \right\rbrace}\,$.
I would like to prove that $F + M$ is dense in $H$, and that $F + M \subsetneq H$.
I have tried to do the analysis of this problem with the vector $\sum_{n=1}^{\infty}\frac{1}{n}e_{-n}$
but I don't know how to proceed.
I would appreciate someone to help me.
For showing that $F+M$ is dense you only have to show that any vector orthogonal to both $F$ and $M$ is $0$ and this is easy.
For the second part your idea is good. Let $x=\sum\limits_{n=1}^{\infty} \frac 1 n e_{-n}$. Observe that $e_{-n}+ne_n, n \geq 1$ are orthogonal. This implies that any element of $M$ is of then form $\sum\limits_{n=1}^{\infty} a_n u_n$. Of course, any element of $F$ is of the form $\sum\limits_{n=0}^{\infty} b_n e_{n}$. If $x \in F+M$ then we get $\sum\limits_{n=1}^{\infty} \frac 1 n e_{-n} =\sum\limits_{n=1}^{\infty} a_n u_n+\sum\limits_{n=0}^{\infty} b_n e_{n}$. But then $a_n=\frac 1 n$ for all $n \geq 1$ and $na_n+b_n=0$ for all $n \geq 1$. But this gives $b_n=-na_n=-1$ a contradiction to $\sum b_n^{2} <\infty$.