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?
 A: 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
$$S_x=1-\frac{1}{2 x}-\frac{1}{6 x^2}+\frac{1}{4
   x^3}+O\left(\frac{1}{x^4}\right)$$
A: 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
$$\sum_{k=1}^xk^p \sim \frac{x^{p+1}}{p+1}$$
A: Hint :
Take, $\displaystyle a_{x,n}=\frac{x^2}{x^2+n}$. Then use Cauchy's First limit theorem.
As $a_{x,n} \to 1$ when $x\to \infty$, so $\displaystyle \frac{1}{x}\sum_{n=1}^{x}a_{x,n} \to 1$ when $x\to \infty$.
A: With a Riemannian sum:
$$\sum_{k=1}^n\frac{n}{n^2+k}=n\frac 1{n^2}\sum_{k=1}^n\frac1{1+\dfrac k{n^2}}$$ can be seen as a Riemaniann sum truncated to the $n$ first terms among $n^2$. Then
$$\lim_{n\to\infty}n\int_0^{1/n}\frac{dx}{1+x}=\lim_{n\to\infty}n\log\left(1+\frac1n\right)=1.$$
No quite rigorous, though (because the sum converges to the integral at the same time that we increase $n$).
A: Just using that $\frac{1}{1+\epsilon}=1-\epsilon+O(\epsilon^2)$ and $\sum_{i=1}^n i=\frac{n(n+1)}{2}$
So $$\frac{x}{x^2+c}=\frac{1}{x}\frac{x^2}{x^2+c}=\frac{1}{x}\frac{1}{1+\frac{c}{x^2}}=\frac{1}{x}\left(1-\frac{c}{x^2}+O(\frac{c}{x^4})\right)$$
$$\sum_{c=1}^x\frac{x}{x^2+c}=\frac{1}{x}\sum_{c=1}^x1-\frac{c}{x^2}+O(\frac{c}{x^4})=1-\frac{x (x+1)}{2x^3}+O(\frac{1}{x^2})=1-\frac{1}{2x}+O(\frac{1}{x^2})$$
Note that using $\frac{1}{1+\epsilon}=1+O(\epsilon)$ is in fact enough and simpler here, but you don't get the $\frac{-1}{2x}$, only $O(\frac{1}{x})$
