Bài 107762

Question

a) Tìm $\mathop {\lim }\limits_{n \to +\infty} \frac{\sqrt[]{9n^2+1}-3n-1 }{\sqrt[]{n^2+4n+1}-n } $
b) Tìm $\mathop {\lim }\limits_{n \to +\infty} u_n$, biết $u_n=\frac{1}{1.3}+\frac{1}{2.4}+...+ \frac{1}{n(n+2)} $

score 0 · Answer 1

a) $u_n=\frac{\sqrt[]{9n^2+1}-3n-1 }{\sqrt[]{n^2+4n+1}-n }=\frac{9n^2+1-(3n+1)^2}{n^2+4n+1-n^2}.\frac{\sqrt[] {n^2+4n+1}+n }{\sqrt[]{9n^2+1}+3n+1 }$
          $=\frac{-6n}{4n+1}.\frac{\sqrt[]{n^2+4n+1}+n }{\sqrt[]{9n^2+1}+3n+1}= \dfrac{-6}{4+\frac{1}{n}}. \dfrac{\left ( 1+\frac{4}{n}

+\frac{1}{n^2}\right )+1 }{\sqrt[]{9+\frac{1}{n^2} }+3+\frac{1}{n} }$
Vì $\lim \frac{1}{n^2}=\lim \frac{1}{n}=0 $. Vậy $\mathop {\lim }\limits_{n \to +\infty} u_n=-\frac{1}{4} $

b) Vì $u_n=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{n(n+2)}=\frac{1}{2} \left ( \frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right )$
Nên   $\lim u_n = \lim \frac{3}{4}-\lim \frac{1}{2(n+1)}-\lim \frac{1}{2}.\frac{1}{n+2}=\frac{3}{4}-0-0     $
Vậy $\mathop {\lim }\limits_{n\to +\infty }u_n=\frac{3}{4} $

Bài 107762

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Bài 107762

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