Đáp án:
$\dfrac{1}{2}$
Giải thích các bước giải:
$\lim \left( \dfrac{1}{1.3}+\dfrac{1}{3.5} + ...+\dfrac{1}{(2n-1)(2n+1)}\right)$
Đặt $P = \left( \dfrac{1}{1.3}+\dfrac{1}{3.5} + ...+\dfrac{1}{(2n-1)(2n+1)}\right)$
$→ 2P = 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$
$→ 2P = =1 - \dfrac{1}{2n+1}$
$→ P = \dfrac{2n}{2n+1}$
Vậy $\lim \left( \dfrac{1}{1.3}+\dfrac{1}{3.5} + ...+\dfrac{1}{(2n-1)(2n+1)}\right)=\lim \dfrac{2n}{2n+1}=\dfrac{1}{2}$
BẠN XEM THAM KHẢO NHA!!!
