Is $\underset{i}\varprojlim R/(I+J)^i \cong \underset{i}\varprojlim \left( \underset{j}\varprojlim R/(I^i + J^j)\right)$? I am studying $I$-adic completions of a ring and I was wondering if the following holds for any commutative ring $R$ and ideals $I, J \subset R$

$\underset{i}\varprojlim R/(I+J)^i \cong \underset{i}\varprojlim \left( \underset{j}\varprojlim R/(I^i + J^j)\right)$

I don't know whether this is true or not, but I can't find a proof nor a counterexample of the above. Thank you!
 A: By general properties of limits $\underset{i}\varprojlim \left( \underset{j}\varprojlim R/(I^i + J^j)\right) \cong \underset{(i,j)}\varprojlim R/(I^i+J^j)$ 
The RHS means that we take the inverse limit over the product category $\Bbb N \times \Bbb N$ which is still a directed set with the ordering given by $(i,j) \leq (i',j') \Leftrightarrow i \leq i' \land j \leq j'$
Note that $(\Bbb N, \leq)$ embeds diagonally into $\Bbb N \times \Bbb N$ with the ordering as described aboved. The image of that is a (co)inital subset (See wikipedia and nlab) because for every element $(i,j) \in \Bbb N \times \Bbb N$, we have $(\min(i,j),\min(i,j)) \leq (i,j)$. General properties of limits and final functors (see the linked nlab page) tell us that we have
$\underset{(i,j)} \varprojlim R/(I^i+J^j) \cong \underset{i}\varprojlim R/(I^i+J^i)$
So we are left to compare $\underset{i}\varprojlim R/(I^i+J^i)$ and $\underset{i}\varprojlim R/(I+J)^i$
Note that $(I+J)^i \supset I^i + J^i$, so we have a projections $R/(I^i+J^i) \to  R/(I+J)^i$. Thus we get a compatible family of morphisms given by the compositions $\underset{i}\varprojlim R/(I^i+J^i) \to R/(I^i+J^i) \to R/(I+J)^i$
(I'll leave out the verification of compatibility)
This induces a morphism $f:\underset{i}\varprojlim R/(I^i+J^i) \to \underset{i}\varprojlim R/(I+J)^i$
On the other hand, we have $(I+J)^{2i} \subset I^i+J^i$, which gives a projection $R/(I+J)^{2i} \to R/(I^i+J^i)$ and then using the same approach, the universal property of the limit gives a morphsim $g:\underset{i}\varprojlim R/(I+J)^i \to \underset{i}\varprojlim R/(I^i+J^i)$
It remains to show that $f$ and $g$ are inverses of each other.
Denote by $\pi_i$ the structure morphism $\underset{i}\varprojlim R/(I^i+J^i) \to R/(I^i+J^i)$ and by $\psi_i$ the structure morphism $\underset{i}\varprojlim R/(I+J)^i \to R/(I+J)^i$ Denote by $p_i$ the canonical projection $R/(I^i+J^i) \to R/(I+J)^i$ and by $q_i$ the projection $R/(I+J)^{2i} \to R/(I^i+J^i)$.
By construction $f$ is the unique morphism such that $\psi_i \circ f = p_i \circ \pi_i$ for all $i$. Similarly, we get that $g$ is the unique morphism such that $\pi_i \circ g= q_i \circ \psi_{2i}$ for all $i$.
This implies that for all $i$ we have $\psi_i \circ f \circ g = p_i \circ \pi_i \circ g= p_i \circ q_i \circ \psi_{2i}$
But $p_i \circ q_i: R/(I+J)^{2i} \to R/(I+J)^i$ is just the standard transition map in the inverse system $(R/(I+J)^{i})_{i \in \Bbb N}$ so that we have $p_i \circ q_i \circ \psi_{2i}=\psi_i$ by compatibility of the structure maps $\psi_i$
Thus we get that $\psi_i \circ f \circ g = \psi_i$ for all $i$. But the universal property of $\underset{i}\varprojlim R/(I+J)^i$ applied to the structure maps $\psi_i$ tells us that there is a unique morphism $h:\underset{i}\varprojlim R/(I+J)^i \to \underset{i}\varprojlim R/(I+J)^i$ satisfying $\psi_i \circ h=\psi_i$. Evidently, the identity is such a morphism. Therefore $f \circ g = \mathrm{id}$
For the other direction, we get for all $i$: $\pi_i \circ g \circ f = 
q_i \circ \psi_{2i} \circ f = q_i \circ p_{2i} \circ \pi_{2i}$.
Note that $q_i \circ p_{2i}:R/(I^{2i}+J^{2i}) \to R/(I^i+J^i)$ is the canonical surjection from the inclusion $I^{2i}+J^{2i} \subset I^i+J^i$ which is also the transition map in the inverse system $(R/(I^i+J^{i}))_{i \in \Bbb N}$, so that we get $q_i \circ p_{2i} \circ \pi_{2i}= \pi_i$ by compatability of the structure maps. So $g \circ f$ satisfies $\pi_i \circ g \circ f= \pi_i$ for all $i$. Thus $g \circ f = \mathrm{id}$ by the same uniqueness argument as for the other direction.
