If a sequence $\{x_n\}$ converges to l, then the sequence $\{y_n\}$ also converges to l, where $y_n=x_1+x_2+\cdots +\dfrac{x_n}{n}$.
I suspect that $\lim_{n\to\infty}\sum_{k=1}^n \dfrac{2k-1}{n^2}$ is 2, using Cauchy theorem, but the answer is 1. Please correct me if I'm wrong.
HINT
Note that
$$\sum_{k=1}^n \frac{2k-1}{n^2}=\frac2{n^2}\sum_{k=1}^n k-\frac1{n^2}\sum_{k=1}^n 1=\frac2{n^2}\frac{n(n+1)}{2}-\frac1n$$
The first claim is true for
$$y_n=\frac{x_1+x_2+⋯+x_n}{n}\to l$$
by Stolz-Cesaro.