A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)
A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)
Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)
nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)
\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)
A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)
A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)
\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\)
Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A> \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)
Tương tự ta có :
A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)
A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)
A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)
\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)
