Consider a collection of $C^\*$-algebras $\{A\_i\}\_{i \in I}$. We can form the direct sum $$\bigoplus\_{i \in I}^{c\_0} A\_i:= \left\{(a\_i)\_{i \in I} \in \prod\_{i\in I} A\_i: \lim\_{i \in I} \|a\_i\| = 0\right\}$$
which is an ideal in the $C^\*$-algebra
$$\bigoplus\_{i \in I}^{\ell^\infty} A\_i:= \left\{(a\_i)\_{i \in I} \in \prod\_{i\in I} A\_i: \sup\_{i \in I} \|a\_i\| <\infty\right\}.$$
Given a $C^\*$-algebra $A$, denote its multiplier algebra by $M(A)$. One possible realisation of the multiplier algebra is by setting $M(A):= \mathcal{L}\_A(A)$, the adjointable operators when we view $A$ as a (right) Hilbert $C^\*$-module over itself.


I believe I have proven that
$$M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) \cong \bigoplus\_{i \in I}^{\ell^\infty} M(A\_i)\cong M\left(\bigoplus\_{i \in I}^{\ell^\infty}A\_i\right)$$
but I can't find a reference for this statement. So, is my assertion true?




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Here is a proof sketch:
We use the implementation of the multiplier $C^\*$-algebra as adjointable operators. We then have natural maps
$$M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) \to \bigoplus\_{i \in I}^{\ell^\infty} M(A\_i): t \mapsto (\iota\_i ^\* t\iota\_i)\_{i \in I}$$ where
$\iota\_i: A\_i \hookrightarrow \bigoplus\_{i \in I}^{c\_0} A\_i$ is the inclusion map and
$$\bigoplus\_{i \in I}^{\ell^\infty} M(A\_i) \to M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) : (t\_i)\_{i \in I} \mapsto [(a\_i)\_i \mapsto (t\_i(a\_i))]$$
These are easily checked to be $\*$-isomorphisms that are inverse to each other, and this establishes the isomorphism $M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) \cong \bigoplus\_{i \in I}^{\ell^\infty} M(A\_i)$. The other isomorphism is shown similarly.



Direct sum of multiplier algebras
Let's try to flesh out your "sketch". Set $A=c\_0-\oplus\_i A\_i$ and consider this as a Hilbert $C^\*$-module over itself. Let $\iota\_i:A\_i\rightarrow A$ be the inclusion, and $\jmath\_i:A\rightarrow A\_i$ the left inverse to $\iota\_i$. These are both non-degenerate $\*$-homomorphisms, and so extend to unital $\*$-homomorphisms $\overline\iota\_i:M(A\_i)\rightarrow M(A)$ and $\overline\jmath\_i:M(A)\rightarrow M(A\_i)$. Also, notice that by definition of the $A$-valued inner-product on $A$,
$$ (b|\iota\_i(a)) = \iota\_i\big( (\jmath\_i(b)|a) \big)
\qquad (b\in A, a\in A\_i). $$


Given $T\in M(A)$ set $T\_i = \overline\jmath\_i(T)\in M(A\_i)$. By definition, $T\_i(\jmath\_i(a)) = \jmath\_i(T(a))$ for $a\in A$, and as $\jmath\_i$ is a $\*$-homomorphism, also $T\_i^\*(\jmath\_i(a)) = \jmath\_i(T^\*(a))$. For $a\in A\_i$ and $b=(b\_j)\in A$,
$$ (T\iota\_i(a)|b) = (\iota\_i(a)|T^\*(b))
= \iota\_i\big( (a|\jmath\_i(T^\*(b))) \big)
= \iota\_i\big( (a|T\_i^\*(\jmath\_i(b))) \big) $$
while
$$ (\iota\_i T\_i(a)|b) = \iota\_i\big( (T\_i(a)|\jmath\_i(b)) \big)
= \iota\_i\big( (a|T\_i^\*(\jmath\_i(b))) \big). $$
Thus $T\iota\_i = \iota\_iT\_i$ for each $i$.


As the linear span of the images of the $\iota\_i$ are dense in $A$, it now follows that
$$ T(a) = \sum\_i T\iota\_i\jmath\_i(a) = \sum\_i \iota\_i\big( T\_i\jmath\_i(a) \big), $$
and the isomorphism $M(A) \cong \ell^\infty-\oplus\_i M(A\_i)$ now follows.




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I actually find arguing using Hilbert $^\*$-modules a bit cumbersome. Instead, let $A\_i\subseteq\mathcal B(H\_i)$ non-degenerately, for each $i$, for some suitable Hilbert space $H\_i$. Set $H = \oplus\_i H\_i$, so naturally $A$ acts non-degenerately on $H$. Set $B=\ell^\infty-\oplus\_i A\_i$, so also $B$ acts non-degenerately on $H$. I'll now consider $B$, but much the same argument works for $A$.


We know that $M(B)\subseteq B''\subseteq \mathcal B(H)$, and indeed
$$ M(B) = \big\{ T\in B'' : Tb, bT\in B \ (b\in B) \big\}. $$
Given $b=(b\_i)\in B$ and $\xi=(\xi\_i)\in H$, by definition, $b(\xi) = (b\_i(\xi\_i))$. As such, with $p\_i\in\mathcal B(H)$ the projection onto the factor $H\_i$, we see that $b p\_i = p\_i b$ so $p\_i\in B'$.


Thus, any $T\in M(B)\subseteq B''$ commutes with each $p\_i$. By linearity and continuity, there is $(T\_i) \in \ell^\infty-\oplus\_i \mathcal B(H\_i)$ with $T(\xi) = (T\_i(\xi\_i))$ for each $\xi\in H$. Using the inclusions $A\_i\rightarrow B$, we can now show that each $T\_i\in M(A\_i)$.


We have hence showed that $M(B) \cong \ell^\infty-\oplus\_i M(A\_i)$. [This seems surprising to me, but I believe this 2nd argument.]


