Consider the series $$\lim \limits_{n \to \infty}\sum_{i=1}^n \frac{n}{n^2+i}$$
Considering individual terms, we get $$\lim \limits_{n \to \infty}\frac{n}{n^2+i}=\lim \limits_{n \to \infty}\frac{1}{n+\frac{i}{n}}=0.$$ Hence the sum would be $0$.
However, considering an alternative approach, we observe that $\frac{n}{n^2+n}<\frac{n}{n^2+i}<\frac{n}{n^2+1}$. Therefore $$\sum_{i=1}^n \frac{n}{n^2+n}<\sum_{i=1}^n \frac{n}{n^2+i}<\sum_{i=1}^n \frac{n}{n^2+1} \implies \frac{n^2}{n^2+n}<\sum_{i=1}^n \frac{n}{n^2+i}<\frac{n^2}{n^2+1}$$ Taking limit $\lim \limits_{n \to \infty}$ on the three expressions, we find that $\lim \limits_{n \to \infty}\frac{n^2}{n^2+n}=\lim \limits_{n \to \infty}\frac{n^2}{n^2+1}=1.$
Therefore by the sandwich theorem, $\lim \limits_{n \to \infty}\sum_{i=1}^n \frac{n}{n^2+i}=1$
Since there exists a unique limit, there must be only one solution. Which approach is the correct one?
There was one suggested reason that each of the individual limits that was calculated in the 1st approach had to be summed up to infinite terms. Since each of the limits were infintesimally small, we cannot predict the way they will behave when summed to infinity and therefore the approach is incorrect. However, my counter was that the limits are exact and are not infinitesimals. Therefore 0 can be summed up infinitely many times and the answer will still be zero.
Is this reasoning correct/wrong? What must be the answer to the problem?
