The limit of a sequence

Given

${x}_{n}=\frac{1}{{n}^{2}+1}+\frac{1}{{n}^{2}+2}+\frac{1}{{n}^{2}+3}+\cdots +\frac{1}{{n}^{2}+n}$

Verify if there is or no a limit. Find it if affirmative.

Let ${a}_{n}=\frac{n}{{n}^{2}+1}$ (The biggest portion of the sum $n$ times) and ${b}_{n}=\frac{n}{{n}^{2}+n}$ (the smallest portion of the sum $n$ times) then

${b}_{n}\le {x}_{n}\le {a}_{n}$

${b}_{n}\le {x}_{n}\le {a}_{n}$

since

$lim\frac{n}{{n}^{2}+1}=lim\frac{n}{{n}^{2}+n}=0,$

we have that

$lim{x}_{n}=0.$

Is this wrong? why? If it is, any tips on how to find $lim{x}_{n}$? Grateful for any help.

Given

${x}_{n}=\frac{1}{{n}^{2}+1}+\frac{1}{{n}^{2}+2}+\frac{1}{{n}^{2}+3}+\cdots +\frac{1}{{n}^{2}+n}$

Verify if there is or no a limit. Find it if affirmative.

Let ${a}_{n}=\frac{n}{{n}^{2}+1}$ (The biggest portion of the sum $n$ times) and ${b}_{n}=\frac{n}{{n}^{2}+n}$ (the smallest portion of the sum $n$ times) then

${b}_{n}\le {x}_{n}\le {a}_{n}$

${b}_{n}\le {x}_{n}\le {a}_{n}$

since

$lim\frac{n}{{n}^{2}+1}=lim\frac{n}{{n}^{2}+n}=0,$

we have that

$lim{x}_{n}=0.$

Is this wrong? why? If it is, any tips on how to find $lim{x}_{n}$? Grateful for any help.