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$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
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.
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 $1$!)
With integrals: $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)=\lim_{n\rightarrow\infty} \frac{1}{n}\biggl(\frac{1}{n} + \frac{2}{n} + \frac{3}{n}+\dots+\frac{n}{n}\biggr)= \int_{0}^1 x dx =\frac{1}{2}$.
Notice that:
$$\lim_{n\rightarrow\infty}\dfrac{1}{n^2} + \dfrac{2}{n^2} + \dfrac{3}{n^2}+\cdots+\dfrac{n}{n^2}=\lim_{n\rightarrow\infty} \frac{1}{n^2}\left(1+2+\dots n\right)$$ and this last sum can be replaced by the Gauss formula, so it becomes: $$\lim_{n\rightarrow\infty}\frac{1}{n^2}\left(\frac{n(n+1)}{2}\right)$$
$$=\lim_{n\rightarrow\infty}\frac{n+1}{2n}=\frac{1}{2}\left(\lim_{n\rightarrow\infty} \frac{n+1}{n}\right)=\frac{1}{2}\left(\lim_{n\rightarrow\infty} 1+\frac{1}{n}\right)=\frac{1}{2}$$
Perhaps this is beyond your curriculum, but one could also use the Stolz-Cesàro theorem.
$$\lim_{n\to\infty}\frac{1+2+\cdots+(n-1)+n}{n^2}$$
Denote the numerator and denominator, respectively, by
$$a_n=\sum_{k=1}^nk$$ $$b_n=n^2$$
It's easy to show that $b_n$ is strictly monotone and divergent. Stolz-Cesàro then says that if
$$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
exists, then it is equal to
$$\lim_{n\to\infty}\frac{a_n}{b_n}$$
We have
$$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\sum\limits_{k=1}^{n+1}k-\sum\limits_{k=1}^nk}{(n+1)^2-n^2}=\lim_{n\to\infty}\frac{n+1}{2n+1}=\frac12$$
As the Wiki page mentions, the theorem is sometimes referred to as "L'Hopital's rule for sequences". If you find that intuitive (related question), then perhaps you will think the same of the approach using this theorem.
If you pull out the denominator, the numerator sums to $\frac{n(n+1)}{2}$, then $n^2$ term cancels out and when you take the limit you get $\frac{1}{2}$
You have asked where you are mistaken, so I will only answer this. You have tried to apply the limit individually to the summands. There is a rule which allows you to do this for finite sums, but here we do not have a fixed number of summands, and this rule does not apply. For a simpler example why you also cannot expect it to apply, consider $$ \lim_{n\to\infty}\Bigl(\underbrace{\frac1n+\frac1n+\dots+\frac1n}_{\text{$n$ summands}}\Bigr). $$ You could again reason that each summand goes to zero, and therefore so must the sum. But of course this is just a fancy way of writing $$\lim_{n\to\infty} 1,$$ and the limit is $1$. So you see that in general you cannot apply the limits individually to the summands of a sum if the number of summands is not fixed and finite.
Indeed by adding zeros on the right you can view the sum as a series. A series is just a limit of partial sums, and so this also gives you a nice example that you cannot in general exchange two limits.
Your intuition "the denominator of every term grows much faster than the numerator" is partially true. It would be completely true if there were finite number of terms. However, you must bear in mind that there are infinite number of terms. Hence you have two variables: denominator $n^2$ and numerator $1+2+\cdots+n$ (when you add up the numerators of the fractions with common denominator). Dealing with infinitely small or large quantities is frequently risky and counterintuitive, because in the given limit you have so called indetermined form of $\frac{\infty}{\infty}$ or $0\cdot \infty$ (if expressed as $\frac{1}{n^2}\cdot (1+2+\cdots+n)$). Thus the result of such indefinite forms depends on their relative rates of increase and/or decrease. As an example, consider a rectangle of area $A$. If its width is reduced infinitely while its length is held constant, then its area will decrease to $0$. If its width is reduced while its length is extended at the same rate, then its area will stay constant. This is what happenned to your limit. And if its width is reduced while its length is extended at a relatively higher rate, the its area will increase infinitely.
Keep developing your intuition!
Seeing such a term, my intuition would be:
The second line is the important one; this immediately tells me that I cannot trust any further intuition (i.e., random ideas shooting spontaneously through my mind) anymore, and that it is time to find a pen and some paper.
Full disclosure: I am more at home in CS than maths, but the same is true for O notation in CS, it can be quite misleading and unintuitive. Or rather, deceivingly simple and provoking intuitive errors.
"grows much faster than the numerator": you are disregarding the fact that the numerator is actually
$$1+2+3+\cdots n, $$ which doesn't grow slower than $n^2$.
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 $S_n\to{1\over2}$.
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.
