# help with $\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right)$

Is my solution right?

\begin{align} \lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right) &=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(1+2+3+\cdots+n-1\right)\right)\\ &=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(\frac{(n-1)(1+n-1)}{2}\right)\right)\\ &=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(\frac{n(n-1)}{2}\right)\right)\\ &=\lim_{n\to\infty}\left(\frac{n-1}{2n}\right)\\ &=\lim_{n\to\infty}\left(\frac{1}{2}\right)\\ &=\boxed{\frac{1}{2}} \end{align}

• Yeah. absolutely. Oct 20, 2021 at 9:34
• You could say first that you are using the known formula $1+2+\cdots +n-1=n(n-1)/2$. Then the second step can be omitted - which is better. This $1+n-1$ is a bit distracting anyway. Oct 20, 2021 at 9:36
• thank you my friend Oct 20, 2021 at 9:37
• I would skip that $\lim_{n\to \infty} \left(\frac{1}{2}\right)$ bit too.
– Gary
Oct 20, 2021 at 9:41
• @DanielG. You know that you have to '@' the user you are sending the comment in order to notify them right? Oct 20, 2021 at 9:54

Correct solution. Another way is to use Riemann sums: \begin{align*} \sum_{k=1}^{n-1}\frac k{n^2} &=\frac 1n\sum_{k=1}^n\frac kn-\frac1n \end{align*} and\begin{align*} \lim_n\left(\frac 1n\sum_{k=1}^n\frac kn\right)=\int_0^1x\;dx=\frac12. \end{align*}
It is absolutely correct. Although while writing the second line, I think it would be better to write $$1+2+3+\cdots+\color{red}{(}n-1\color{red}{)}$$ instead of $$1+2+3+\cdots+n-1$$ as it may be interpreted as the sum to $$n$$ and then subtracting a $$1$$.