Ta luôn có
$\dfrac{n}{\sqrt{n^2+n}}\le A=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}} \le \dfrac{n}{\sqrt{n^2+1}}$
Mà $\lim \dfrac{n}{\sqrt{n^2+n}} =\lim \dfrac{n}{\sqrt{n^2+1}} =1 \Rightarrow \lim A=1$
b) Chứng minh quy nạp được
$0 <B= \frac{1}{2}.\frac{3}{4}...\frac{2n-1}{2n} < \dfrac{1}{\sqrt{3n+4}}$
Mà $\lim \dfrac{1}{\sqrt{3n+4}}=0 \Rightarrow \lim B=0$