$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
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.
Best Answer
Intuition should say:
the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.
And that's where intuition stops. From here on, you go with logic and rigor, not intuition.
And it gets you to
$$\frac1{n^2} + \frac2{n^2}+\cdots + \frac{n}{n^2} = \frac{1+2+3+\cdots + n}{n^2} = \frac{\frac{n(n+1)}{2}}{n^2} = \frac{n^2+n}{2n^2}$$
and you find that the limit is $\frac12$ (not $1$!)