Direct sum of multiplier algebras 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?

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.
 A: 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.

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.]
