<s>[INST] An interesting limit: $\lim_{n\rightarrow \infty}\frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\cdots+\frac{n}{n^{2}+n}$ $$\lim_{n\rightarrow \infty}\frac{1}{n^{2}＋1}＋\frac{2}{n^{2}＋2}＋…＋\frac{n}{n^{2}＋n}$$
Thanks in advance
 [/INST] HINT:
$$\frac{1}{n^2+n}\sum_{k=1}^n(k)\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{1}{n^2+1}\sum_{k=1}^n (k)$$
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We can write the limit as$$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k}$$Then, using $n^2+n\ge n^2+k\ge n^2+1$ we have $$\frac{1}{n^2+n}\sum_{k=1}^n(k)\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{1}{n^2+1}\sum_{k=1}^n (k)\tag 1$$We can sum the  arithmetic progression to obtain $$\sum_{k=1}^n (k)=\frac{n(n+1)}{2}\tag 2$$Using $(2)$ in $(1)$ reveals $$\frac{n(n+1)}{2(n^2+n)}\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{n(n+1)}{2(n^2+1)}\tag 3$$whereupon applying the squeeze theorem to $(3)$ yields the coveted limit $$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k}=\frac12$$ 

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