Question

chứng minh rằng : C=1/2.3/4.5/6. ... .9999/10000<1/100

Answer 1

Ta cóC= \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}......\dfrac{9999}{10000}\)

Đặt A = \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{10000}{10001}\)

Khi đó AC = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}\)= \(\dfrac{1}{10001}\)

Do \(\dfrac{1}{2}< \dfrac{2}{3}\)

\(\dfrac{3}{4}< \dfrac{4}{5}\)

.............

\(\dfrac{9999}{10000}< \dfrac{10000}{10001}\)

=> C<A=>C²<CA hay C²< \(\dfrac{1}{10001}\) , mà \(\dfrac{1}{10001}\)<\(\dfrac{1}{10000}\)=> C²< \(\dfrac{1}{10000}\)

^{Khi đó C < \(\sqrt{\dfrac{1}{10000}}\)}hay C< \(\dfrac{1}{100}\)( đpcm )