TITLE: help with $\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right)$
QUESTION [5 upvotes]: 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}

REPLY [2 votes]: 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*}