calculus - Evaluating the limit $lim_{ntoinfty} left(frac{1}{n^2}+frac{2}{n^2}+...+frac{n-1}{n^2}right)$

Evaluate the limit $$\lim_{n\to\infty} \left(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right) .$$

My work:

I started by computing the first 8 terms of the sequence $x_n$ ($0, 0.25, 0.222, 0.1875, 0.16, 0.139, 0.122, 0.1094$). From this, I determined that the sequence $x_n$ monotonically decreases to zero as n approaches infinity. which satisfies my first test for series convergance, if $\sum_{n=2}^\infty x_n$ converges, then $\lim_{n\to\infty}x_n=0$.

Next, I rearranged the equation in an attempt to perform a comparison test. I re-wrote the equation as $\sum_{n=2}^\infty (\frac1{n}-\frac1{n^2})$. This was to no avail as the only series I could think to compare it to was $\frac1n$ which is always greater than the original series and is divergant, which does not prove convergance to a limit.

The ratio test concluded with $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}$ being equal to 1, which is also inconclusive (I can show my work if need be, but that would be a little tedious). I never ran the root test, but I doubt that this would be any help in this case.

I see no other way to compute the limit, so any help would be appreciated!!

Answer

Hint: Use the formula
$$\sum_{k=1}^n k=\frac{n(n+1)}2.$$

So,
$$\lim_{n\to\infty}\frac 1{n^2}+\frac 2{n^2}+\cdots+\frac{n-1}{n^2}=\lim_{n\to\infty}\frac{n(n-1)}{2n^2}.$$

Can you get it from here?