<s>[INST] What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$? 
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$

I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.
Where is my mistake? Please explain very intuitively.
 [/INST] Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero.  An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have
$$
\frac{1}{4} = \frac{n}{2}\cdot \frac{n/2}{n^2} \le \frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n-1}{n^2} + \frac{n}{n^2} \le n\cdot \frac{n}{n^2} = 1
$$
In fact, the answer is just the limit of $n^{-2}\binom{n+1}{2} = \frac{1}{2}$ as $n\to\infty$, but the above is why you should guess that the answer isn't zero.
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