<s>[INST] The limit of a sum: $\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$ 
Evaluate the following limit:
  $$
\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
$$

I haven't ever taken the limit of the sum... Where do I start?
Do I start taking the sum?
 [/INST] With Riemann sums:
We have
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}-\left(\frac{k}{n}\right)^2\right)
= \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\left(1-\frac{k}{n}\right)
$$
which is a Riemann sum for $f\colon[0,1]\to\mathbb{R}$ defined by $f(x)=x(1-x)$. Therefore, we have
$$
\lim_{n\to\infty} \sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) = \int_0^1 f(x)dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 =\boxed{ \frac{1}{6}}\,.
$$
Without Riemann sums:
Here, you can directly use the facts that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ to compute the sum, and conclude afterwards.
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n^2}\sum_{k=1}^n k-\frac{1}{n^3}\sum_{k=1}^n k^2 = \frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{6n^3} \xrightarrow[n\to\infty]{} \frac{1}{2} - \frac{1}{3} = \boxed{\frac{1}{6}}\,.
$$
</s>