Does $\frac{1}{n}\sum_{k=0}^{n-1}f(k)=f(n)+\epsilon_n$ imply that $f(n)$ converges? If we let $f(n)$ be an arbitrary arithmetic function with the property that $\forall$ $n>0$
$$\frac{1}{n}\sum_{k=0}^{n-1}f(k)=f(n)+\epsilon_n$$
where $\lim_{n\to\infty}\epsilon_n=0$, then does $\lim_{n\to\infty}f(n)$ converge? I know that if $\lim_{n\to\infty}f(n)$ converges then the equation holds, but the converse seems much trickier since $f(n)$ isn't necessarily bounded. I suspect that this equivalence holds but I simply do not know how to prove it.
 A: By evaluating your expression at $n=1,2,\ldots$ you can obtain a formula for $f(n)$ in terms of $f(0)$. I get
$$
f(1)= f(0)-\epsilon_1$$
$$
f(2) = f(0)-\frac12\epsilon_1-\epsilon_2
$$
$$
f(3) = f(0)-\frac12\epsilon_1-\frac13\epsilon_2-\epsilon_3
$$
$$
f(4) = f(0)-\frac12\epsilon_1-\frac13\epsilon_2-\frac14\epsilon_3-\epsilon_4
$$
and in general (setting $\epsilon_0:=0$)
$$
f(n)=f(0) -\epsilon_n-\sum_{k=1}^n\frac{\epsilon_{k-1}}k,$$
which you can prove by induction$^{(*)}$.
With this representation we see that $f(n)$ need not converge; for example take $\displaystyle\epsilon_k=\frac1{\log (k+1)}$.

(*) The base case $n=1$ is clear. If the claim holds for $n$, then
$$
\begin{aligned}
f(n+1)+\epsilon_{n+1}&=\frac1{n+1}\sum_0^n f(k)\\
&=\frac1{n+1}\sum_{k=0}^n\left( f(0) - \epsilon_k-\sum_{j=1}^k\frac{\epsilon_{j-1}}j\right)\\
&=f(0)-\frac1{n+1}\sum_{k=1}^{n+1}\epsilon_{k-1}-\frac1{n+1}\sum_{k=0}^n\sum_{j=1}^k\frac{\epsilon_{j-1}}j.
\end{aligned}$$
To evaluate that final term, interchange the order of summation to get
$$
\begin{aligned}
\frac1{n+1}\sum_{j=1}^n\sum_{k=j}^n\frac{\epsilon_{j-1}}j
&=\frac1{n+1}\sum_{j=1}^n\frac{n+1-j}j\epsilon_{j-1}\\
&=\sum_{j=1}^n\frac{\epsilon_{j-1}}j-\frac1{n+1}\sum_{j=1}^n\epsilon_{j-1}\\
&=\sum_{j=1}^{n+1}\frac{\epsilon_{j-1}}j-\frac1{n+1}\sum_{j=1}^{n+1}\epsilon_{j-1}.
\end{aligned}
$$
Putting it all together, this yields the claim for $n+1$.
