Let $H=(H,(\cdot, \cdot))$ be a Hilbert space and $(u_n)_{n \in \mathbb{N}} \subset H$ a complete orthonormal set in $H$. Given $u \in H$, in some papers, for instance $[1, \text{Theorem 4.10}]$ and $[3, \text{page 257}]$, the authors consider a series of the form $$ \sum_{n\in \mathbb{N}\setminus J}\frac{1}{a_n}(u, v_n)v_n \cdot a_n \tag{1} $$ where $a_n \in \mathbb{R}$, for all $n \in \mathbb{N}$, and $J:=\{i \in \mathbb{N} \; ; \; a_i=0\}$ is a finite set. Obviously, by $(1)$ we have $$ \sum_{n \in \mathbb{N}\setminus J}\frac{1}{a_n}(u, v_n)v_n \cdot a_n=\sum_{n \in \mathbb{N}\setminus J}(u, v_n)v_n. $$
On the other hand, it is well know (see $[2$, Theorem $9.12$ $]$) that we can write $$ u=\sum_{n \in \mathbb{N}}(u, v_n)v_n = \sum_{n \in \mathbb{N}\setminus J}(u, v_n)v_n+\sum_{n \in J}(u, v_n)v_n. \tag{2} $$
Making a notational convention, namely $$ 0=\frac{0}{0} \tag{3} $$ the authors write then by $(1)$ and $(2)$ the following $$ \sum_{n\in \mathbb{N}\setminus J}\frac{1}{a_n}(u, v_n)v_n \cdot a_n =\sum_{n\in \mathbb{N}}\frac{1}{a_n}(u, v_n)v_n \cdot a_n = u. $$
Question. In this context, why the notational convention in $(3)$ can be considered?
In the book $[4, \text{Section $8.5$}]$ the division by zero is discussed, but I haven't been able to conclude whether or not I can even consider $(3)$. Any suggestion?
$[1]$ Albert, John P. Positivity Properties and Stability of Solitary–Wave Solutions of Model Equations For Long Waves. Communications in partial differential equations $17.1-2 (1992): 1-22.$
$[2]$ Bachman G. and Narici, L., Functional Analysis. New York: Academic Press, $2000$.
$[3]$ Freeden, Willi, and M. Zuhair Nashed. Ill-posed problems: operator methodologies of resolution and regularization. Handbook of Mathematical Geodesy. Birkhäuser, Cham, $2018. 201-314.$
$[4]$ Suppes, P., Introduction to logic. D. Van Nostrand Co., Inc., New York, $1957$.
There is no serious mathematical content behind that convention... and certainly no assertion that $0/0=0$ in any numerical sense. Rather, to say this is a (rather cavalier) version of saying that, if $a_n=0$, then just drop that term entirely in the renormalized infinite sum.
The only small mathematical point is that the formula for properly renormalizing the coefficients is not correct for $a_n=0$... For other reasons, we know that in that case the $n$th renormalized coefficient should still be $0$...
I myself would indeed be disquieted by saying that a thing is true "by convention", as opposed to giving a good reason. In that regard, the author somewhat misrepresented the situation, but there's nothing serious hiding behind that glibness.