Erdős Problem 897: an explicit slowly growing counterexample
Statement and result.
An additive function means a function on positive integers with $f(ab)=f(a)+f(b)$ when $\gcd(a,b)=1$. We construct a positive such function, for $n\geq2$, satisfying \[
\frac{f(p^k)}{\log(p^k)}\longrightarrow\infty
\quad(p^k\longrightarrow\infty),
\qquad
\frac{f(n+1)-f(n)}{\log n}\longrightarrow0,
\qquad
\frac{f(n+1)}{f(n)}\longrightarrow1.
\] Thus both proposed infinite-limsup conclusions are false. The construction makes precise the slowly growing weights credited to Erdős and discussed by Wirsing; no novelty is claimed.
A slowly varying weight on prime powers.
For real $x>0$, let $L(x)$ be the least nonnegative integer $j$ such that applying the natural logarithm $j$ times to $x$ gives a number at most one. In particular, $L(x)=0$ for $x\leq1$, and \[
L(x)=1+L(\log x)\quad(x>1).
\] Set $g(x)=\sqrt{L(x)+1}$. This is nondecreasing, unbounded, and $g(x)=O(\sqrt{\log x})$ as $x\to\infty$. For all sufficiently large $x$, if $t=\sqrt{\log x}$, then $t\geq\log\log x$, whence \[
L(t)\geq L(x)-2,\qquad
0\leq g(x)-g(t)
\leq\sqrt{L(x)+1}-\sqrt{L(x)-1}\longrightarrow0.
\tag{1}
\] The jumps of $L$ occur at the iterated-exponential thresholds and these thresholds eventually have gaps greater than one. Hence $L(n+1)-L(n)\in\{0,1\}$ for all sufficiently large integers $n$, and $g(n+1)-g(n)\to0$.
Definition and a uniform approximation.
Put $f(1)=0$, and, using the exact prime-power factors of $n$, define \[
f(n)=\sum_{p^k\parallel n}g(p^k)\log(p^k).
\] Unique prime factorization proves additivity on coprime arguments. Also $f(p^k)/\log(p^k)=g(p^k)\to\infty$. We claim \[
f(n)=g(n)\log n+o(\log n).
\tag{2}
\] Indeed $\sum_{p^k\parallel n}\log(p^k)=\log n$. Split this sum at $t=\sqrt{\log n}$. The factors exceeding $t$ contribute at most $(g(n)-g(t))\log n=o(\log n)$ to the nonnegative error $g(n)\log n-f(n)$, by $1$. There are at most $\lfloor t\rfloor$ remaining factors, each with logarithm at most $\log t$. Their contribution is at most $g(n)t\log t=o(\log n)$. For the last estimate one may use $L(n)=1+L(\log n)=O(\log\log n)$, which follows from $L(y)=O(\log y)$. Thus the error bound is uniform over all factorizations of $n$, proving $2$.
Consecutive differences and ratios.
The main term in $2$ satisfies \[
\frac{g(n+1)\log(n+1)-g(n)\log n}{\log n}
=g(n+1)-g(n)
+g(n+1)\frac{\log(1+1/n)}{\log n}\longrightarrow0.
\] Subtracting $2$ at $n$ and $n+1$ gives the asserted limit zero for the normalized difference. Furthermore $f(n)/\log n=g(n)+o(1)\to\infty$, so \[
\frac{f(n+1)}{f(n)}
=1+\frac{f(n+1)-f(n)}{f(n)}\longrightarrow1.
\] The ratio assertion concerns $n\geq2$, where the denominator is positive; $f(1)=0$ causes no limiting difficulty.
Sources and scope.
Current statement and historical attribution:
https://www.erdosproblems.com/897, checked 24 September 2026. All estimates needed for this counterexample were proved above. It addresses ordinary additive functions; it does not assert the same result under the extra restrictions of strong or complete additivity suggested in the historical discussion. No local Lean verification is claimed.
Completion estimate.
COMPLETION: 100%