Calculate $\lim_{n \to \infty} \left(\frac{n}{n^2 + 1} + \frac{n}{n^2 + 2} + \cdots + \frac{n}{n^2 + n}\right)$ 
Calculate $$\lim_{n \to \infty} \bigg(\frac{n}{n^2 + 1} + \frac{n}{n^2 + 2} + \cdots + \frac{n}{n^2 + n}\bigg)$$

Hello!
The answer given to this problem is $1$, but I am getting to $0$:
Consider $$\lim_{n \to \infty} \frac{n}{n^2 + k} \stackrel{\text{L'Hopital}}{=} \lim_{n \to \infty} \frac{1}{2n} = 0.$$
Now, since in the original expression, all the individual limits exist and are defined, the answer is $0 + 0 + \cdots + 0 = 0$, which is wrong answer.
I can guess where I go wrong: the number of terms is something that depends on $n$, but when doing L'Hopital, I am treating is as constant.
But, I am still not clear about this. How exactly is this method wrong?
Please note that my question is not about how to solve this question, but about why a particular method is wrong. I already know it has got answers in a different post.
 A: For each term, $\frac{n}{n^2+k}$ it goes to arbitrarily small, say $\frac{n}{n^2+k}\to \epsilon$. But you have infinitely many terms sum together, so what is the sum of infinitely many $\epsilon$? It is undetermined!
Note that:
$$\frac{n}{n^2+n}\le\frac{n}{n^2+k}\le\frac{n}{n^2+1}$$
So we have:
$$\frac{n^2}{n^2+n}=\sum_{k=1}^n\frac{n}{n^2+n}\le\sum_{k=1}^n\frac{n}{n^2+k}\le\sum_{k=1}^n\frac{n}{n^2+1}=\frac{n^2}{n^2+1}$$
Further:
$$1=\lim_{n\to\infty}\frac{n^2}{n^2+n}\le\lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k}\le\lim_{n\to\infty}\frac{n^2}{n^2+1}=1$$
From Squeeze theorem, we know the limit is $1$
A: Summarizing comments to an answer as the OP requested:
The claim

If $A_i:=\lim_{n\to\infty}a_i(n)$ exist for $1\le i\le k$ then $\lim_{n\to\infty}\sum_{i=1}^ka_i(n)=\sum_{i=1}^kA_i$

is famous in the case $k=2$, from which we can prove it for all finite integers $k\ge0$ by induction on $k$. This does not, however, prove an infinite-$k$ or $n$-dependent $k$ case. Since @MathFail's answer shows the true limit is $1$, this problem provides a counterexample to any such conjecture.
