I was wondering whether the following assertion is true.
Given the series $\sum_{i=1}^{n-1} {a_i\frac{i}{n}}$ with $\sum_{-\infty}^{\infty}|a_i|<\infty$
Is that enough to guarantee that $\lim_{n \to \infty}\sum_{i=1}^{n-1} {a_i\frac{i}{n}}=0$?
I was trying to prove it but didn't get anywhere. I am starting to think that the assertion is in fact false. Does anyone know a counterexample?
The assertion is true. Given $\varepsilon > 0$, by the assumed absolute convergence there is an $m \in \mathbb{N}$ such that
$$\sum_{i = m+1}^{\infty} \lvert a_i\rvert < \frac{\varepsilon}{2}\,.$$
Then for such a fixed $m$, there is an $N \in \mathbb{N}$ such that
$$\frac{1}{n} \sum_{i = 1}^m \lvert a_i\rvert\cdot i < \frac{\varepsilon}{2}$$
for all $n \geqslant N$. For $n > \max \{m, N\}$ we therefore have
\begin{align} \Biggl\lvert \sum_{i = 1}^{n-1} a_i\frac{i}{n}\Biggr\rvert &\leqslant \Biggl\lvert \sum_{i = 1}^m a_i \frac{i}{n}\Biggr\rvert + \Biggl\lvert \sum_{i = m+1}^{n-1} a_i \frac{i}{n}\Biggr\rvert \\ &\leqslant \sum_{i = 1}^m \lvert a_i\rvert \frac{i}{n} + \sum_{i = m+1}^{n-1} \lvert a_i\rvert \frac{i}{n} \\ &\leqslant \frac{1}{n}\sum_{i = 1}^m i\lvert a_i\rvert + \sum_{i = m+1}^{n-1} \lvert a_i\rvert \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}\,. \end{align}