The group of additive sequences of integers

This is a proof which is completely elementary (although not too simple). The idea is to prove that, for $n$ sufficiently big, $f(e_{n})=0$.

Consider a sequence of the kind $x=(2^{n_{1}},2^{n_{2}},...)$, where $(n_{i})$ is an increasing sequence of natural numbers (I'll explain later how I'm going to choose that sequence). For every $k$, we have

$$ f(x)=f(2^{n_{1}},...,2^{n_{k}},2^{n_{k+1}},...)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+f(0,...,0,2^{n_{k+1}},2^{n_{k+2}},...)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+2^{n_{k+1}}f(0,...,0,1,2^{n_{k+2}-n_{k+1}},2^{n_{k+3}-n_{k+1}},...). $$

Call $b_{k}=f(0,...,0,1,2^{n_{k+2}-n_{k+1}},2^{n_{k+3}-n_{k+1}},...)$, so that

$$f(x)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+2^{n_{k+1}}b_{k}.$$

Therefore,

$$b_{k}=\dfrac{1}{2^{n_{k+1}}}\left[f(x)-\sum_{i=1}^{k}2^{n_{i}}f(e_{i})\right].$$

Now, applying the Triangle Inequality,

$$|b_{k}|\leq \dfrac{1}{2^{n_{k+1}}}\left[|f(x)|+\sum_{i=1}^{k}2^{n_{i}}|f(e_{i})|\right].$$

By choosing $(n_{i})$ appropriately, we can make the right hand side of the above inequality tend to $1/2$. Simply choose by recursion the sequence so that for every $k$,

$$\sum_{i=1}^{k}2^{n_{i}}|f(e_{i})|<\dfrac{1}{2}2^{n_{k+1}},$$

and we have convergence to $1/2$. Since $b_{k}$ is an integer for every $k$, we have $b_{k}=0$ for $k\geq N$ ($N$ being a sufficiently large natural number).

Now, simply notice that $b_{k}=f(e_{k+1})+2^{n_{k+2}-n_{k+1}}b_{k+1}$ (this follows directly from the definition of $b_{k}$). Using this, we deduce immediately that $f(e_{k})=0$ for large $k$.

Hope this helps!

So this all appears in Section 94 of Fuchs's Infinite Abelian Groups II, and this is due to Sasiada. The general result is that this property holds whenever the image $G$ is countable and reduced (i.e., has no non-trivial divisible subgroup, where a divisible group is one where, given any $x\in G$ and $n\in \mathbb N$, there exists $y\in G$ such that $x=n\cdot y$). I will write the proof in the case of $G=\mathbb Z$, but it is a simple matter to generalize this.

Let $P$ be the product of countably many copies of $\mathbb{Z}$ and $G=\mathbb{Z}$. Let $\eta:P\to G$ be a homomorphism such that $\eta(e_n)\neq 0$ for infinitely many $n$. Of course, by removing those for which $\eta(e_n)=0$, we may assume that $\eta(e_n)$ is non-zero for all $n$. Since $\bigcap_m mG=0$, there exists a sequence of integers $$ 1=k_1<k_2<\cdots$$ such that $\eta(2k_n!e_n)\not\in k_{n+1}G$ for each $n=1,2,\dots$. The set of elements $(g_1,g_2,\dots)$ in $P$ with each $g_i$ equal to either $0$ or $k_i!$ has cardinality $2^{\aleph_0}$, and $G$ has strictly smaller cardinality. Thus there there are two different elements of this set that have the same image under $\eta$. Their difference is an element $x=(h_1,h_2,\dots)$ with each $h_i$ either $0$, $\pm k_i!$ or $\pm 2k_i!$.

But now we obtain a contradiction. If $j$ is the first index for which $h_j\neq 0$ (and there must be such an index as $x\neq 0$) then $\eta(h_j)\not\in k_{j+1}G$, and $$ \eta(h_j)=\eta(x)-\eta(0,\dots,0,h_{j+1},h_{j+2},\dots)=-\eta(0,\dots,0,h_{j+1},\dots)\in k_{j+1}G.$$

(There appears to be a typo in the proof of 94.2 in Fuchs. He only requires that $\eta(k_i!e_i)\not\in k_{i+1}G$ for all $i$, whereas it seems to me that you need the factor of $2$ because it looks like he gets the possibilities for the entries of $x$ wrong, and omits the potential $2k_i!$, and only has $0$ or $\pm k_i!$ as options for $h_i$.)