Vtyjkyuk

Find $$lim_xto inftyleft(x over x^2+1+x over x^2+2+cdots +xover x^2+xright)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let left[left(x over x^2+1+x over x^2+2+cdots +xover x^2+xright)right]=f(x)implies \ left(x over x^2+x+x over x^2+x+cdots +xover x^2+xright)lt f(x)lt left(x over x^2+1+x over x^2+1+cdots +xover x^2+1right) \ x^2 over x+x^2lt f(x) lt x^2over 1+x^2\ textapplying limit on both sides \ implieslim_xto inftyx^2 over x+x^2= lim_xto inftyx^2over 1+x^2=1\ implies lim_xto inftyf(x)=1$$
Can we do this without squeeze theorem?

We could do it using harmonic numbers
$$S_x=sum_k=1^x frac x x^2+k=xsum_k=1^x frac 1 x^2+k=x left(H_x^2+x-H_x^2right)$$
Now, using the asymptotics
$$H_p=gamma +log left(pright)+frac12 p-frac112
p^2+Oleft(frac1p^4right)$$ we should get
$$S_x=1-frac12 x-frac16 x^2+frac14
x^3+Oleft(frac1x^4right)$$

By geometric series we have

$$frac x x^2+k=frac1xfrac 1 1+k/x^2=$$$$=frac1xleft(1-frackx^2+left(-frackx^2right)^2+ldotsright)=frac1x-frackx^3+frack^2x^5+ldots$$

and therefore

$$sum_k=1^x frac x x^2+k=sum_k=1^x left(frac1x-frackx^3+frack^2x^5+ldotsright)=1-fracsum_k=1^xkx^3+fracsum_k=1^xk^2x^5+ldots $$

$$sim 1-fracx^22x^3+fracx^33x^5+ldotsto 1$$

indeed recall that by Faulhaber's formula

$$sum_k=1^xk^p sim fracx^p+1p+1$$

Hint :

Take, $displaystyle a_x,n=fracx^2x^2+n$. Then use Cauchy's First limit theorem.

As $a_x,n to 1$ when $xto infty$, so $displaystyle frac1xsum_n=1^xa_x,n to 1$ when $xto infty$.

With a Riemannian sum:

$$sum_k=1^nfracnn^2+k=nfrac 1n^2sum_k=1^nfrac11+dfrac kn^2$$ can be seen as a Riemaniann sum truncated to the $n$ first terms among $n^2$. Then

$$lim_ntoinftynint_0^1/nfracdx1+x=lim_ntoinftynlogleft(1+frac1nright)=1.$$

No quite rigorous, though (because the sum converges to the integral at the same time that we increase $n$).

Just using that $frac11+epsilon=1-epsilon+O(epsilon^2)$ and $sum_i=1^n i=fracn(n+1)2$

So $$fracxx^2+c=frac1xfracx^2x^2+c=frac1xfrac11+fraccx^2=frac1xleft(1-fraccx^2+O(fraccx^4)right)$$

$$sum_c=1^xfracxx^2+c=frac1xsum_c=1^x1-fraccx^2+O(fraccx^4)=1-fracx (x+1)2x^3+O(frac1x^2)=1-frac12x+O(frac1x^2)$$

Note that using $frac11+epsilon=1+O(epsilon)$ is in fact enough and simpler here, but you don't get the $frac-12x$, only $O(frac1x)$