Solution: et $k \in \mathbb{N} $. We will find $\left\vert \left\langle g_j, f_k \right\rangle \right\vert^2 $ for $j \in \mathbb{N} $. \begin{align} \left\vert \left\langle g_j, f_k \right\rangle \right\vert^2 & = \left\vert \left\langle \sum_{n=1}^{j} (-1)^{n+1} e_i, e_k + e_{k+1} \right\rangle \right\vert ^2 \notag\\[1ex] & = \left\vert \sum_{i=1}^{j} (-1)^{n+1} \left\langle e_i, e_k + e_{k+1} \right\rangle \right\vert ^2 \label{eq:06Jan2024-1} \end{align} Note that if $j < k$, then \[ \left\vert \left\langle g_j, f_k \right\rangle \right\vert = 0 \implies \left\vert \left\langle g_j, f_k \right\rangle \right\vert^2 = 0. \] If $j = k$, then using \eqref{eq:06Jan2024-1} and the previous observation, \begin{align*} \left\vert \left\langle g_j, f_k \right\rangle \right\vert^2 & = \left\vert (-1)^k \left\langle e_k, e_k + e_{k+1} \right\rangle \right\vert ^2 \\ & = \left\vert (-1)^k\right\vert = 1. \end{align*} Similarly, if $j \geq k + 1$, then \begin{align*} \left\vert \left\langle g_j, f_{k+1} \right\rangle \right\vert^2 & = \left\vert (-1)^k \left\langle e_k, e_k + e_{k+1} \right\rangle + (-1)^{k+1} \left\langle e_{k+1}, e_k + e_{k+1} \right\rangle \right\vert ^2 \\ & = \left\vert (-1)^k + (-1)^{k+1} \right\vert = 0. \end{align*} Thus, we obtained \begin{equation}\label{eq:06Jan2024-2} \left\vert \left\langle g_j, f_k \right\rangle \right\vert ^2 = \begin{cases} 0, &\text{ if } j \neq k ;\\ 1, &\text{ if } j = k. \end{cases} \end{equation}

Now using \eqref{eq:06Jan2024-2} we have \begin{align*} \sum_{k=1}^{\infty} \left\vert \left\langle g_j, f_k \right\rangle \right\vert ^2 = \left\vert \left\langle g_j, f_j \right\rangle \right\vert ^2 = 1. \end{align*} Thus, the correct answer will be option (D).