Em tham khảo:
Ta có
$\dfrac{1}{3^2}>$$\dfrac{1}{3}-$$\dfrac{1}{4}$
$\dfrac{1}{4^2}>$$\dfrac{1}{4.5}=$$\dfrac{1}{4}-$$\dfrac{1}{5}$
$\dfrac{1}{5^2}>$$\dfrac{1}{5.6}=$$\dfrac{1}{5}-$$\dfrac{1}{6}$
$.....$
$\dfrac{1}{50^2}>$$\dfrac{1}{50}-$$\dfrac{1}{51}$
⇒$A>$$\dfrac{1}{3}-$$\dfrac{1}{4}+$$\dfrac{1}{4}-$$\dfrac{1}{5}+$$\dfrac{1}{5}-$$\dfrac{1}{6}+......$$\dfrac{1}{50}-$$\dfrac{1}{51}$
⇒$A>$$\dfrac{1}{3}-$$\dfrac{1}{51}=$$\dfrac{16}{51}>$$\dfrac{1}{4}$
Tương tự
$A<$$\dfrac{1}{2}-$$\dfrac{1}{3}+$$\dfrac{1}{3}-$$\dfrac{1}{4}+....$$\dfrac{1}{49}$-$\dfrac{1}{50}$
⇔$A<$$\dfrac{4}{9}<$$\dfrac{12}{25}$
Vậy $\dfrac{1}{4}<A<$$\dfrac{4}{9}$
Học tốt
