Prove that $\sum\limits_{n=1}^{\infty}a_n< \infty $ imply $\lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} n a_n=0$ Let $ \{a_n\} \ n \in \mathbb{N} $ a sequence of non-negative real numbers. 
Prove that
$\displaystyle\sum_{n=1}^{\infty}a_n< \infty $ imply $\displaystyle\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n \leq N} n a_n=0$
Would this be accomplished without the condition $a_n \geq 0$?
 A: You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $g\in L^1_+[0,\infty)$. Then $$\frac 1n \int_0^n xg(x)dx\to 0$$ as $n\to\infty$.
Proof Consider $g_n(x)= \dfrac{xg(x)}n {\bf 1}_{[0,n]}$. Then $|g_n|\leqslant |g|$ and $g_n(x)\to 0$. By DCT, we conclude.
A: Let $A_n = a_1 + \cdots + a_n$ be the partial sum of $\{a_n\}$, by summation by parts formula
$$\sum_{n = 1}^N na_n = \sum_{n = 1}^{N - 1}A_n(n - (n + 1)) + NA_N = NA_N - \sum_{n = 1}^{N - 1}A_n.$$
Thus, based on the famous Cesaro's theorem
$$\lim_{N \to \infty} \frac{1}{N}\sum_{n = 1}^N na_n = \lim_{N \to \infty} A_N - \lim_{N \to \infty} \frac{\sum_{n = 1}^{N - 1}A_n}{N} = a - a = 0$$
where $a = \sum_{n = 1}^\infty a_n < \infty$. 
The above proof shows that as long as $\left|\sum_{n = 1}^\infty a_n\right| < \infty$, the result always holds. $a_n$ could be negative.
A: $\dfrac{\displaystyle \sum_{n=1}^N na_n}{N}=\dfrac{a_1+2a_2+3a_3+\cdots +Na_N}{N}=\dfrac{a_1+\cdots+a_N}{N}+\dfrac{N-1}{N}\cdot\dfrac{a_2+\cdots+a_N}{N-1}+\cdots + \dfrac{a_N}{N}\to 0+0+\cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N \to 0$ when $N \to \infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n \geq 0$.
