limits - Sum of $lim_{nrightarrow infty}left(frac{n}{n^2+1}+frac{n}{n^2+2}+cdots cdots cdots +frac{n}{n^2+n}right)$

Sum of

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

$\bf{My\; Try::}$ I have solved it using Squeeze theorem

$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+n}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+1}$$

So we get

$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r} = 1$$

My question is can we solve above limit without using Squeeze Theorem,

If yes then plz explian me, Thanks

Answer

Using harmonic numbers

$$S_n=\sum^{n}_{r=1}\frac{n}{n^2+r}=n \left(H_{n^2+n}-H_{n^2}\right)$$ Now, using the expansion $$H_p=\gamma+\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p ^4}\right)$$ and applying to each term you should arrive to $$S_n=n\left(\frac{-6 n^3-6 n^2+2 n+1}{12 n^4 (n+1)^2}+\log \left(1+\frac{1}{n}\right)+O\left(\frac{1}{n ^2}\right)\right)$$ Taylor again, $$S_n=1-\frac{1}{2 n}+O\left(\frac{1}{n ^2}\right)$$

Edit

May be of interest

$$T_n=\sum^{n}_{r=1}\frac{n}{n^k+r}=n \left(H_{n^k+n}-H_{n^k}\right)$$ leads to $$T_n=\frac{ \left(n^{2-k}-6 n^{k+1}-6 n^2+2 n\right)}{12 n^k\left(n^k+n\right)^2}+\log \left(1+\frac 1 {n^{k-1}}\right)$$ which makes $$n^{k-1}T_n=1-\frac{1}{2 n^{k-1}}+O\left(\frac{1}{n^k}\right)$$