Giải thích các bước giải:
$A=\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+...+\dfrac{1}{\sqrt{n^2+n}}$
Ta có :
$A<\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2}}+...+\dfrac{1}{\sqrt{n^2}}=\dfrac{n}{\sqrt{n^2}}=1$
$A>\dfrac{1}{\sqrt{n^2+n}}+\dfrac{1}{\sqrt{n^2+n}}+...+\dfrac{1}{\sqrt{n^2+n}}=\dfrac{n}{\sqrt{n^2+n}}$
$\to \lim \dfrac{n}{\sqrt{n^2+n}}<\lim A<\lim 1$
$\to 1<\lim A< 1$
$\to \lim A=1$
$\to \lim \dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+...+\dfrac{1}{\sqrt{n^2+n}}=1$
