When and why is the sum of infinite zeroes exactly zero?

Question

I was solving the following limit problem :

$$L=\lim_{n\to\infty}\sum_{r=1}^{n}\frac{r}{n^2+r^2}$$

Upon opening sum, we have :

$$L= \frac{1}{n^2+1}+\frac{2}{n^2+4}...+\frac{n}{n^2+n}$$

Clearly, each term approaches to zero as $n\to\infty$ . But , my teacher said we cannot add all zeroes and say that answer is zero. This is because there are infinitely many zeroes and $(0\times\infty$) is indeterminate. However, in various limits problems where we had to use series expansions ( like of $\lim_{x\to0}\frac{e^x-1}{x} , \frac{\sin(x)}{x}$, etc.) , we did put zeroes and ignored the sum of infinite zeroes.

For example :

$$\frac{\sin(x)}{x} = 1- \frac{x^2}{6}+\frac{x^4}{120}...$$ As $x\to0$ ,

$$\lim_{x\to0}\frac{\sin(x)}{x}=1+0 +0+0+0...$$ Similarly, for any $n≤r≤1$,

$$\lim_{n\to\infty}\frac{r}{n^2+r^2}=0 $$ So that ; $$\lim_{n\to\infty}\sum\frac{r}{n^2+r^2}=0 + 0 + 0... $$

We get the correct answer in first case but it's wrong in second.

Question:

When and why can the sum of infinite zeroes be exact zero or indeterminate ?

Edit:

I am not interested in finding the answer to this limit problem. My question is about when and when not can we add infinitely many terms approaching zero. The limit used above in my question was just an example to illustrate my main question.

Answer 1

Your example can be written as $$ L = \lim_{n\to\infty} \sum_{r=1}^\infty f_r(n) $$ with $f_r(n) = \frac{r}{n^2+r^2} \,\mathbf 1_{n\geq r}$, and your question is a subcase if the question: when can we interchange limits and infinite sums, i.e. when is it true that $$ \lim_{n\to\infty} \sum_{r=1}^\infty f_r(n) = \sum_{r=1}^\infty \lim_{n\to\infty} f_r(n). $$ A usual sufficient criterion is uniform convergence of the partial sums, which can be written $\sum_{r=R}^\infty f_r(n)\underset{R\to\infty}\to 0$ uniformly with respect to $n$. Another (weaker) one is the dominated convergence theorem which tells you that it will work if there is a summable sequence $v_r$ independent of $n$ such that $|f_r(n)|\leq v_r$. You can check that these criteria work for the Taylor series you mention (convergence is uniform in $x\in[0,C]$ for any fixed constant $C$) but not for $f_r(n)$, mainly because of the $1/r$ behavior. More precisely, if you try for example to use the dominated convergence criterium, then you will want to bound $$ \sup_n |f_r(n)| = \frac{1}{2r} $$ and this is indeed not summable. At the contrary, you can see that if it was instead $f_r(n) = \frac{1}{n^2+r^2} \,\mathbf 1_{n\geq r}$, that would be summable uniformly in $n$ since smaller than $1/r^2$ and so the limit of the sum would indeed be $0$.