Let $(a_n), (b_n)$ be sequences in a Banach algebra and $(c_n) := (a_n.b_n)$
If $\lim (c_n) = c$ and $\lim (a_n) = a $ does it follow that $(b_n)$ has a limit $b$ satisfying $c = a.b$ ?
I can see this result if the sequence $(a_n)$ is invertible, but this isn't guaranteed in a general Banach algebra. Some sort of proof by contradiction seems possible ?
I know I shouldn't really change the question, but rather than posting a new one let me ask in the light of feedback...
Is this true if $a \ne 0$ ?
Note: Let $B^\times$ be the set of invertible elements of $B$. If $a$ is invertible then because $B^\times$ is an open set, the tail of the sequence is invertible. Furthermore the inverse function $\phi: B^\times \to B^\times: \phi(x) = x^{-1}$ is continuous and therefore sequentially continuous.
(Ref: Megginson An Introduction to Banach Space Theory p.310)
Then $\lim a_i = a \implies \lim a_i^{-1} = a^{-1}$ (at least as this applies to the invertible tail of the sequence).
As a standard result on limits $\lim x_i = x$ and $\lim y_i = y$ then $\lim x_i.y_i = x.y$
So, $\lim [(a_i^{-1}).(a_i.b_i)] = \lim b_i = a^{-1}.c$
I.e. the result holds if $a$ is invertible
This isn't even true in $\mathbb{C}$. Take $a_n=c_n=0$ and $b_n=(-1)^n$.
Edit: If you want $a\ne 0$ then consider $A=M_2(\mathbb{C})$ and the following sequences: $a_n=\begin{pmatrix}1&0 \\ 0 & \frac{1}{n} \end{pmatrix}, b_n=\begin{pmatrix}1&0 \\ 0 & (-1)^n \end{pmatrix}, c_n=\begin{pmatrix}1&0 \\ 0 & \frac{(-1)^n}{n} \end{pmatrix}$. Note that here the elements of $a_n$ are even invertible.