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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.

ejtrsmft
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    As a general principle, these Cesaro summation converses never hold in generality. Best to look for counterexamples. – Joshua P. Swanson Mar 19 '21 at 22:24
  • As you say, if $\lim_x \frac{1}{x} \sum_{n \le x} a_n$ is convergent, then it must converge to $k$. So if you want to look for a counterexample, this must be divergent. – Crostul Mar 19 '21 at 22:45

3 Answers3

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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.

perroquet
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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).

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

reuns
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  • I didn't downvote, but I really don't find this answer clear at all, and as far as I can tell it deserves its current rating. The 10 doesn't do anything, so I'm not sure why it's included. You say, "What do you get with", but it's not clear if you're genuinely asking a question or if it's supposed to be rhetorical and you're trying to give a hint. Most importantly, I'm not convinced this is actually a counterexample. There seems to be a lot of oscillation and I think neither sum converges. – Joshua P. Swanson Mar 20 '21 at 06:59
  • @JoshuaP.Swanson It is a counter-example. $x^{-1}\sum_{n\le x} a_n$ oscillates in $[9-\epsilon,11+\epsilon]$ while $\frac1{\log x} \sum_{n\le x} a_n/n \to 10$, the latter is because $\frac{(n+1)^{1+i}-n^{1+i}}{n} = (n+1)^i-n^i+n^{i-1}+O(n^{-2})$ – reuns Mar 20 '21 at 18:07