$$\lim_{n\to\infty}{ \frac {n^2}{1+2+…+n}}$$
Here, if we divide and multiply by $n^2$, the numerator becomes $1$ and the limit of the denominator becomes $0$ then, the entire limit becomes infinite. But if I actually do the summation of the numbers in the denominator and then divide by $n^2$ I get a finite answer $2$… I can’t understand why this happens?
On
As notice in the comments and in the given fine answer the denominator divided by $n^2$ tends to $\frac12 $ and not to zero. The key fact is that in the limit we are summing "an infinite number of terms" (that is a larger and larger number of terms) and for that reason their sum can be finite, even if each term tends to zero.
As an alternative, we can also obtain the result by Cesaro-Stolz theorem
$$\frac {(n+1)^2-n^2}{(1+2+\ldots+n+n+1)-(1+2+\ldots+n)}=\frac{2n+1}{n+1} \to 2$$
You say that $$\lim_{n\to \infty}\frac{1}{1/n^2+2/n^2+\cdots+1/n}=\infty, $$ since the denominator goes to zero, which is not true because the number of terms involved in the sum of the denominator changes within the value of $n$, so its not correct. A direct approach using $1+2+\cdots+n=n(n+1)/2$ gives us that $$\lim_{n\to \infty}\frac{n^2}{1+2+\cdots+n}=\lim_{n\to \infty}\frac{2n^2}{n^2+n}=2.$$