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
 A: Added for your curiosity.
Sooner or later, you will learn about harmonic numbers and it will be clear to you that $$S_n=\sum_{i=1}^n\frac{i}{n^2+i}=n \left(n H_{n^2}-n H_{n^2+n}+1\right)\tag 1$$ For large values of $p$, the asymptotics is $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)\tag 2$$ Applying $(2)$ in $(1)$, you will get as an approximation $$S_n=\frac{1}{12} \left(-\frac{1}{n^2}+12 n+\frac{6 n+7}{(n+1)^2}\right)+n^2 \log \left(\frac{n^2}{n^2+n}\right)+\cdots$$ Using Taylor series for the logarithm and long division for the remaining term, you should end with $$S_n=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+\frac{2}{15
   n^3}+O\left(\frac{1}{n^4}\right)\tag 3$$ which, for sure, shows the limit and how it is approached. It also gives a very simple approximation of the result.
For example, using $n=10$, the exact result is $$S_{10}=\frac{3039003639041255}{5909102214621606}\approx 0.5142919396$$ while $(3)$ would give $\frac{5143}{10000}$.
A: HINT:
$$\frac{1}{n^2+n}\sum_{k=1}^n(k)\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{1}{n^2+1}\sum_{k=1}^n (k)$$
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

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 $$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k}=\frac12$$ 

