Converse to a result of convergence of $\lim\frac{1}{x}\sum_{n\leq x}a_n=k$

Question

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.

Answer 1

What do you get with $$a_n = 10+ \Re( (n+1)^{1+i}-n^{1+i})$$

Answer 2

Without loss of generality, we set $k=1$ and define

$$ A(x)=\sum_{1\le n\le x}{a_n\over n}=\log x+R(x) $$

wherein $R(x)=o(\log x)$. Then, it follows from partial summation that

$$ \begin{aligned} \sum_{n\le x}a_n &=\int_{1^-}^xt\mathrm dA(t)=xA(x)-\int_1^xA(t)\mathrm dt \\ &=x\log x+xR(x)-\int_1^x\log t\mathrm dt-\int_1^xR(t)\mathrm dt \\ &=x\log x-xR(x)-[t\log t-t]_1^x-\int_1^xR(t)\mathrm dt \\ &=x-1-xR(x)-\int_1^xR(t)\mathrm dt \end{aligned} $$

Certainly, more constraints are needed onto $R(x)$ in order for us to reduce the remaining terms (or otherwise we can deduce PNT directly from Mertens' first theorem).

Answer 3

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.