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.

Here's an explanation based on the intuitively obvious fact that a triangle with very little height can nonetheless have substantial area provided it has a very long base.
Draw the step function $f_n(x)={\lceil x\rceil\over n^2}$ for $0\le x\le n$, and observe that it lies above the triangle with vertices at $(0,0)$, $(n,0)$ and $(n,{1\over n})$ -- i.e., a right triangle with base of length $n$ and height $1\over n$, hence area $1\over2$.  Since the area beneath the step function is $S_n={1\over n^2}+{2\over n^2}+\cdots+{n\over n^2}$, this shows that $\lim_{n\to\infty}S_n$, if it exists, cannot be less than $1\over2$.  
If you like, you can also draw the step function $g_n(x)={\lfloor x\rfloor\over n^2}$ for $0\le x\le n+1$ which lies below a right triangle with base of length $n+1$ and height $n+1\over n^2$ and argue that the area beneath the step function, ${0\over n^2}+{1\over n^2}+{2\over n^2}+\cdots+{n\over n^2}$, is less than ${1\over2}\left(n+1\over n\right)^2$.  Combining this with the result of the first paragraph leads to the limit $S_n\to{1\over2}$.  
