Let $p$ be prime, $R$ be a ring and $M_n$ be a $R$-modules such that $p^nM_n = 0, p^{n-1}M_n \neq 0$. Set $M' := \text{Hom}_R(\oplus_{n>0}M_n, \oplus_{n>0}M_n)$.
Is $M'$ torsion? I.e. is $M' \otimes \mathbb Q = 0$?
I'm confused because of the following: take $f = id \in M'$. Then there is no $t \in \mathbb N$ such that $p^tf = 0$. Yet, for any $t \in \mathbb N$, $f \otimes 1 = f \otimes \frac{p^t}{p^t} = p^tf \otimes \frac{1}{p^t} \in M' \otimes \mathbb Q$. So if we want to evaluate $f\otimes 1$ at, say, $m:=(m_1,\cdots,m_n,0,\cdots)$ where $m_i \in M_i$, we get $$ (f\otimes 1)(m) = (p^nf\otimes \frac{1}{p^n})(m) = f(p^n m) \otimes \frac{1}{p^n} = 0. $$
Let $M = \bigoplus_n M_n$. You only prove that the image of $$\hom(M,M) \otimes \mathbb{Q} \to \hom(M \otimes \mathbb{Q},M \otimes \mathbb{Q})$$ is zero (which is because we have $M \otimes \mathbb{Q}=0$), which is not enough to conclude that $\hom(M,M) \otimes \mathbb{Q}$ is zero.
In fact, if $\hom(M,M) \otimes \mathbb{Q}$ is zero, we must have $z \cdot \mathrm{id}_M = 0$ for some $z \in \mathbb{Z} \setminus \{0\}$, which means that $z \cdot M_n = 0$ for all $n$. By assumption, this means that $p^n \mid z$ for all $n$, which implies $z=0$, a contradiction.