Question

b, Cho tổng : \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\) Chứng minh \(\dfrac{3}{5}< S< \dfrac{4}{5}\)

Answer 1

\(\dfrac{1}{31}>\dfrac{1}{40}\)

\(\dfrac{1}{32}>\dfrac{1}{40}\)

...

\(\dfrac{1}{40}=\dfrac{1}{40}\)

=>\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\)

\(\dfrac{1}{41}>\dfrac{1}{50}\)

\(\dfrac{1}{42}>\dfrac{1}{50}\)

...

\(\dfrac{1}{50}=\dfrac{1}{50}\)

=>\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\)

\(\dfrac{1}{51}>\dfrac{1}{60}\)

\(\dfrac{1}{52}>\dfrac{1}{60}\)

...

\(\dfrac{1}{60}=\dfrac{1}{60}\)

=>\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{10}{60}=\dfrac{1}{6}\)

=>\(S>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{3}{5}\)

\(\dfrac{1}{31}< \dfrac{1}{30}\)

\(\dfrac{1}{32}< \dfrac{1}{30}\)

...

\(\dfrac{1}{40}< \dfrac{1}{30}\)

=>\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)

\(\dfrac{1}{41}< \dfrac{1}{40}\)

\(\dfrac{1}{42}< \dfrac{1}{40}\)

...

\(\dfrac{1}{50}< \dfrac{1}{40}\)

=>\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\)

\(\dfrac{1}{51}< \dfrac{1}{50}\)

\(\dfrac{1}{52}< \dfrac{1}{50}\)

...

\(\dfrac{1}{60}< \dfrac{1}{50}\)

=>\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\)

=>\(S< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{4}{5}\)

=>\(\dfrac{3}{5}< S< \dfrac{4}{5}\)