If $ \sum\frac{a_n}{n}$ converges, then $\frac{a_1+\cdots+a_n}{n}$ converges to 0? I'm studying Strong Law of Large Numbers proof, and I think above statement is a non-trivial one, so I want to know whether the above statement is true or not.
Thanks for the help and suggestions.
 A: The statement is true. This is Kronecker's lemma.
If $(x_n)_{n=1}^\infty$ is an infinite sequence of real numbers such that 
$$
\sum_{m=1}^\infty x_m = s
$$ 
exists and is finite, then we have for all $0<b_1 \leq b_2 \leq b_3 \leq \ldots$ and $b_n \to \infty$ that 
$$
\lim_{n \to \infty}\frac1{b_n}\sum_{k=1}^n b_kx_k = 0.
$$
A: Let $\sum_{n=1}^\infty \frac{a_n}{n} = a > 0$, and let $\mu$ be a natural number. Then if $\varepsilon > 0$, there exists $N\in\Bbb{N}$ so that $m>N$ implies $|a-\sum_{n=1}^m\frac{a_n}{n}|< \varepsilon/4$. Then if $M > m > N$, you can show
$$|\sum_{n=m+1}^M \frac{a_n}{n}| < \varepsilon/2 $$
Now we have, for each $M=\mu m+k$ (where $0\leq k < \mu$ and $m>N$), 
$$|A_M| = |\frac{a_1+a_2+...+a_M}{M}| \leq |\frac{a_1+a_2+...+a_m}{M}|+|\frac{a_{m+1}+...+a_M}{M}| \leq $$
$$\frac{1}{\mu}|\sum_{n=1}^m\frac{a_n}{n}| + |\sum_{n=m+1}^M \frac{a_n}{n}| < \frac{a+\varepsilon/4}{\mu}+\varepsilon/2 < \frac{a}{\mu}+\varepsilon$$
Since this works for all $\varepsilon>0$, we have that $\lim_{M\to\infty} |A_M| \leq \frac{a}{\mu}$ for all $\mu \in \Bbb{N}$, so $\lim_{M\to\infty} A_M = 0$.
EDIT: This only works if the $a_i$ eventually all have the same sign. See comments. 
