Converse to a result of convergence of $\lim\frac{1}{x}\sum_{n\leq x}a_n=k$ In this post it is proved that, if $(a_n)$ is a sequence in $\mathbb C$:
$$
\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}a_n=k\implies\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k.
$$
Is the converse to this statement true? In other words, is there a sequence $(a_n)$ with
$$
\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k
$$
but $\lim_{x\to\infty}\frac1x\sum_{n\leq x}a_n$ converges to some other limit, or diverges? Thanks.
If $a_n=(-1)^n n$, then:
$\displaystyle \lim_{x \rightarrow +\infty} \dfrac{1}{\log x} \sum_{n\leqslant x} \dfrac{a_n}{n} = \lim_{x \rightarrow +\infty} \dfrac{1}{\log x} \sum_{n\leqslant x} (-1)^n = 0 $
$\displaystyle \lim_{x \rightarrow +\infty} \dfrac{1}{ x} \sum_{n\leqslant x}a_n$ doesn't exist.
What do you get with $$a_n = 10+ \Re( (n+1)^{1+i}-n^{1+i})$$