<s>[INST] Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x  \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x  \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\  \left({x \over x^2+x}+{x  \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x  \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$ 
Can we do this without squeeze theorem?
 [/INST] We could do it using harmonic numbers
$$S_x=\sum_{k=1}^x \frac x {x^2+k}=x\sum_{k=1}^x \frac 1 {x^2+k}=x \left(H_{x^2+x}-H_{x^2}\right)$$
Now, using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ we should get
$$S_x=1-\frac{1}{2 x}-\frac{1}{6 x^2}+\frac{1}{4
   x^3}+O\left(\frac{1}{x^4}\right)$$
</s>