Dùng giới hạn kẹp, làm cho câu khó câu dễ làm theo tương tự $\dfrac{1}{\sqrt{n^2+n}}+\dfrac{1}{\sqrt{n^2+n}}+...+\dfrac{1}{\sqrt{n^2+n}} < \dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+...+\dfrac{1}{\sqrt{n^2+n}}<\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+1}}$
$\Rightarrow \dfrac{n}{\sqrt{n^2+n}}< \dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+...+\dfrac{1}{\sqrt{n^2+n}}<\dfrac{n}{\sqrt{n^2+1}}$
Mà $\lim \dfrac{n}{\sqrt{n^2+n}} =\lim \dfrac{n}{\sqrt{n^2+1}} =1 $
$\Rightarrow\lim \dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+...+\dfrac{1}{\sqrt{n^2+n}}=1$