 # The limit of a sequence Given x_n=(1)/(n^2+1)+(1)/(n^2+2)+(1)/(n^2+3)+...+(1)/(n^2+n) Verify if there is or no a limit. Find it if affirmative. Let a_n=(n)/(n^2+1) (The biggest portion of the sum n times) and b_n=(n)/(n^2+n) (the smallest portion of the sum n times) then b_n<=x_n<=a_n since lim (n)/(n^2+1)=lim (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 limxn? Grateful for any help. Ledexadvanips 2022-08-12 Answered
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.
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Your reasoning is true, but I think the following a bit of better.
$0<{x}_{n}
Thus,
$0\le \underset{n\to +\mathrm{\infty }}{lim}{x}_{n}\le \underset{n\to +\mathrm{\infty }}{lim}\frac{1}{n}=0,$
which says $\underset{n\to +\mathrm{\infty }}{lim}{x}_{n}=0$