Đáp án:
$a)\quad \dfrac12$
$b)\quad 1$
Giải thích các bước giải:
a) Ta có:
$\quad \left(1-\dfrac14\right)\left(1 -\dfrac19\right)\cdots\left(1 -\dfrac{1}{n^2}\right)$
$= \left(1-\dfrac12\right)\left(1 -\dfrac13\right)\cdots\left(1 -\dfrac1n\right)\left(1+\dfrac12\right)\left(1 +\dfrac13\right)\cdots\left(1 +\dfrac1n\right)$
$= \dfrac12\cdot\dfrac23\cdots\dfrac{n-1}{n}\cdot\dfrac32\cdot\dfrac43\cdots\dfrac{n+1}{n}$
$= \dfrac1n\cdot\dfrac{n+1}{2}$
$=\dfrac{n+1}{2n}$
Do đó:
$\quad \lim\left[\left(1-\dfrac14\right)\left(1 -\dfrac19\right)\cdots\left(1 -\dfrac{1}{n^2}\right)\right]$
$=\dfrac12\lim\dfrac{n+1}{n}$
$=\dfrac12\lim\dfrac{1+\dfrac1n}{1}$
$=\dfrac12$
b) Ta có:
$\quad \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots +\dfrac{1}{n(n+1)}$
$=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\cdots +\dfrac{n+1 - n}{n(n+1)}$
$=1 -\dfrac12 +\dfrac12 -\dfrac13 +\dfrac13 -\dfrac14 +\cdots +\dfrac1n -\dfrac{1}{n+1}$
$= 1 -\dfrac{1}{n+1}$
Do đó:
$\quad \lim\left[\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots +\dfrac{1}{n(n+1)}\right]$
$=\lim\left(1 -\dfrac{1}{n+1}\right)$
$= 1 -\lim\dfrac{1}{n+1}$
$= 1 -\lim\dfrac{\dfrac1n}{1+\dfrac1n}$
$= 1$
