# Does $\sum_{i=1}^\infty a_i/i < \infty$ imply that $a_i$ has Cesaro mean zero?

If $(a_i)_{i=1}^\infty$ is a sequence of positive real numbers such that:

$$\sum_{i=1}^\infty \frac{a_i}{i} < \infty.$$

Does this mean that the sequence $(a_i)_{i=1}^\infty$ has Cesaro mean zero? As in

$$\lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n a_i = 0.$$

• I don't think the title of the question agrees with what you ask. Isn't a series Cesaro summable iff average of its partial sums converge? The series $a_i=1/i$ satisfies the equality in the last line you refer in the question but is not Cesaro summable. – Burak Aug 27 '14 at 2:36
• @Burak, the sequence $a_i=1/i$ is Cesaro summable to zero. But I am happy to change the title if you think I could be clearer. – Daniel Mansfield Aug 27 '14 at 2:39
• We must have different notions of Cesaro summation. What I had in mind was this definition: en.wikipedia.org/wiki/Ces%C3%A0ro_summation In this case, the partial sums, which are harmonic numbers, approach to $ln(n)$ for sufficient large n. According to this definition we must take $\frac{1}{n}\Sigma_1^n H_k$, not $\frac{1}{n}\Sigma_1^n \frac{1}{k}$ where $H_k$ is the $k$-th harmonic number. – Burak Aug 27 '14 at 2:44
• @Burak, if I understand you correctly, I should say the sequence of partial sums $s_n = a_1 + \cdots + a_n$ is Cesaro summable; not the sequence $a_n$. – Daniel Mansfield Aug 27 '14 at 2:58
• I guess it's a matter of terminology. It seems that there is a notion of Cesaro mean (en.wikipedia.org/wiki/Ces%C3%A0ro_mean) and Cesaro summation. I think you were referring to the former. It's not that I disagree with the proposition, rather, am confused with the terminology. – Burak Aug 27 '14 at 3:02

Dominated convergence theorem, sum version. Let $b_n$ be the sequence $$b_n(k) = \cases{a_k/n & if k \le n\cr 0 & otherwise\cr}$$ Then $|b_n| \le c$ where $c(k) = a_k/k$, and $b_n \to 0$ pointwise. Since $c \in \ell^1$, we conclude that $\lim_{n \to \infty} \sum b_n = \sum \lim_{n \to \infty} b_n = 0$.
Summation by parts gives: $$\sum_{i=1}^{n}a_i=\sum_{i=1}^{n}\frac{a_i}{i}+\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}\frac{a_k}{k},\tag{1}$$ while the convergence of $\sum_{i=1}^{+\infty}\frac{a_i}{i}$ gives that for any $\varepsilon>0$ there exists $M_\varepsilon$ such that $$\sum_{n\geq M_\varepsilon}\frac{a_i}{i}\leq \varepsilon.$$ Hence, by assuming $\sum_{n=1}^{+\infty}\frac{a_n}{n}=C$, we have, through $(1)$: $$\sum_{i=1}^{KM_\varepsilon}a_i\leq C+CM_\varepsilon+\varepsilon(K-1)M_\varepsilon,\tag{2}$$ hence: $$\limsup_{K\to +\infty}\frac{1}{KM_\varepsilon}\sum_{i=1}^{K M_\varepsilon}a_i \leq \varepsilon.\tag{3}$$ Since $\varepsilon$ was arbitrary, this proves that $\{a_i\}_{i\in\mathbb{N}^*}$ is Césaro summable to zero as conjectured.
Put $S_0=0$ and for $n\geq 1$, $\displaystyle S_n=\sum_{k=1}^n \frac{a_k}{k}$; note that $S_n$ is convergent, say to $S$. We have for $n\geq 1$ that $a_n=n(S_n-S_{n-1})$ and hence $a_1+\cdots+a_n=nS_n-\sum_{k=0}^{n-1} S_k$. Thus $$T_n=\frac{a_1+\cdots +a_n}{n}=S_n-\frac{\sum_{k=1}^{n-1}S_k}{n}$$ and $T_n\to S-S=0$, and we are done.