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
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n a_i = 0.$$
 A: 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 $\lim_{n \to \infty} \sum b_n = \sum \lim_{n \to \infty} b_n = 0$.
A: 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. 
A: 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.
A: I just discovered that this result has a name. If anyone wanted a reference, it is called Kronecker's lemma.
