$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} }$ is dense but not the entire space 
Regarding my previous question: $\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} \cap \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } = 0$, but their sum is dense.

In the context of the above question: $\mathcal{H}$ is a separable Hilbert space, with a total orthonormal set $\{e_n \}_n$. I have able to prove that $$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} \cap \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } $$ is an internal direct sum and that $$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } $$ is dense in $\mathcal{H}$. However, I would now like to prove that $$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } \neq \mathcal{H}. $$
I have tried to do so by considering the sequence $$x_n = \frac{1}{n+1}e_{2n+1} = \left( e_{2n} + \frac{1}{n+1}e_{2n+1} \right) - e_{2n},$$ and at its sequence of partial sums $$s_n = \sum_{i=1}^n x_i. $$
The sequence $(s_n)_n$ does converge in $\mathcal{H}$ as, the sequence $(1/n)_n$ is an element of $\ell^2$. However, is its limit in the direct sum from above?
 A: You are on the right track: No, the limit is not contained in the direct sum. Which proves the question claim.
The proof depends on a fact in Banach space theory:
Let $B$ be a Banach space and consider closed subspaces $X,Z\subset B$. If $B$ is algebraically isomorphic to the direct sum $X\oplus Z$, then the isomorphism is also topological, i.e., an isomorphism of Banach spaces.
This is implied by the Inverse mapping theorem which states "The inverse of a continuous linear bijection between Banach spaces is continuous." (being a corollary to the Open mapping theorem).
Furthermore, the projectors onto the subspaces are then continuous operators on $B$.
Generalising your choice of $(s_n)$, let $\lambda\in\ell^2(\mathbb N)$ satisfy $\lambda_{2n+1}\neq 0\,$ for infinitely many $n$, but $\lambda_{2n}=0$ for all $n$.
This gives
$$v\:=\:\sum_{n=0}^\infty\lambda_{2n+1}e_{2n+1}\in\mathcal{H}$$
and it is shown, after appropriate choice of $\lambda$, that $v\notin F+G\,$ where $F,G$ denote the closed subspaces in question.
Look at the partial sums
$$\begin{align}v_N & \:=\:\sum_{n=0}^N\lambda_{2n+1}e_{2n+1}\:=\:
\sum_{n=0}^N\lambda_{2n+1}\big[(n+1)\big(e_{2n}+\tfrac1{n+1}e_{2n+1}\big)
 -(n+1)e_{2n}\big]\\[1ex]
 & \:=\:\sum_{n=0}^N\lambda_{2n+1}\sqrt{(n+1)^2+1}\; \frac{e_{2n}+\tfrac1{n+1}e_{2n+1}}{\sqrt{1+\frac1{(n+1)^2}}}
 \quad-\quad\sum_{n=0}^N\,(n+1)\lambda_{2n+1}\,e_{2n}\end{align}$$
which are finite linear combinations, thus certainly $v_N\in F+G$.
It is shown in another post that $F\cap G=\{0\}$, so that we really deal with a direct sum  decomposition $F\oplus G$ here. Then the splitting in the last formula line into $F$- and $G$-components is uniquely determined, and these components are the images under the two projections onto $F$ and $G$ of $v_N$.
If $F\oplus G$ were all of $\mathcal H$, then the projections are continuous by the above fact, and the convergence $v_N\rightarrow v$ implies convergence in $F$ and $G$ separately.
Now many possible choices for $\lambda\in\ell^2(\mathbb N)$ carry on sabotage of convergence in $F$ and $G$. Your choice made in the question  is one of them since the second summand then becomes $\sum_{n=0}^N e_{2n}\,$.
This proves $F\oplus G\not= \mathcal H.\qquad\blacksquare$
