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?

By geometric series we have
$$\frac x {x^2+k}=\frac1x\frac 1 {1+k/x^2}=$$$$=\frac1x\left(1-\frac{k}{x^{2}}+\left(-\frac{k}{x^{2}}\right)^2+\ldots\right)=\frac1x-\frac{k}{x^{3}}+\frac{k^2}{x^{5}}+\ldots$$
and therefore
$$\sum_{k=1}^x \frac x {x^2+k}=\sum_{k=1}^x \left(\frac1x-\frac{k}{x^{3}}+\frac{k^2}{x^{5}}+\ldots\right)=1-\frac{\sum_{k=1}^xk}{x^{3}}+\frac{\sum_{k=1}^xk^2}{x^{5}}+\ldots $$
$$\sim 1-\frac{x^2}{2x^{3}}+\frac{x^3}{3x^{5}}+\ldots\to 1$$
indeed recall that by Faulhaber's formula
$$\sum_{k=1}^xk^p \sim \frac{x^{p+1}}{p+1}$$
