Evaluating $\lim _{m\to\infty}\left(\frac1{m^2}+\frac2{m^2}+\frac3{m^2}+\cdots+\frac{m}{m^2}\right)$. Where's my error?

Question

Question is following

$$\lim _{m \to \infty}\left(\frac{1}{m^{2}}+\frac{2}{m^{2}}+\frac{3}{m^{2}}+\cdots+\frac{m}{m^{2}}\right)$$

method-1 $$\lim _{m \rightarrow \infty}\left(\frac{m(m+1)}{2 m^{2}}\right)$$ $$=\frac{1}{2}$$ method-2(applying limits individually)

$$\lim _{m \rightarrow \infty} \frac{1}{m^{2}}+\lim _{m \rightarrow \infty} \frac{2}{m^{2}}+\lim _{m \rightarrow \infty} \frac{3}{m^{2}}+\infty=0$$

Since denominator is always greater than the numerator

What’s wrong with method-2. Is it wrong to apply limits individually. If so then how?

Answer 1

First case is correct. Second case wrong: limit of sums equal sum of limits when we have fixed amount of summands.

Answer 2

The first case works because you have replaced the sum of a varying number of terms with a finite sum of exactly 2 terms, and the replacement is exact with no approximation $$\lim _{m \rightarrow \infty} \left(\frac{1}{2}+\frac{1}{2m} \right)$$ This means it is legitimate to interchange the limit and the sum.

The second case is not correct due to there still being a varying number of terms in the sum. It is not legitimate to apply limits term-by-term in such a case. See the comments for a nice counter-example.