Ta có S = \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)⇒ S < \(\dfrac{1}{30}\cdot10+\dfrac{1}{40}\cdot10+\dfrac{1}{50}\cdot10=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)

Vậy S < \(\dfrac{4}{5}\)

1/31>1/40

1/32>1/40

...

1/40=1/40

=>1/31+1/32+...+1/40>1/40*10=1/4

1/41>1/50

1/42>1/50

...

1/50=1/50

=>1/41+1/42+...+1/50>10/50=1/5

1/51>1/60

1/52>1/60

...

1/60=1/60

=>1/51+1/52+...+1/60>10/60=1/6

=>S>1/4+1/5+1/6=3/5

1/31<1/30

1/32<1/30

...

1/40<1/30

=>1/31+1/32+...+1/40<1/30*10=1/3

1/41<1/40

1/42<1/40

...

1/50<1/40

=>1/41+1/42+...+1/50<10/40=1/4

1/51<1/50

1/52<1/50

...

1/60<1/50

=>1/51+1/52+...+1/60<10/50=1/5

=>S<1/3+1/4+1/5=4/5

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

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

...

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

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

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

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

...

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

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

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

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

...

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

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

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

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

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

...

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

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

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

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

...

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

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

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

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

...

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

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

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

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

S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)

=>S>1/40*10+1/50*10+1/60*10=3/5

S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)

=>S<1/30*10+1/40*10+1/50*10=4/5

=>3/5<S<4/5

Giải:

Đặt \(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)

Ta có:

\(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)

\(\Rightarrow A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)

Nhận xét:

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

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

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

\(\Rightarrow A< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)

\(\Rightarrow A< \dfrac{4}{5}\left(1\right)\)

Lại có:

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

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

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

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\)

\(\Rightarrow A>\dfrac{3}{5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)

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

Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)

Đặt

Ta có:

Nhận xét:

Lại có:

Từ và

Vậy

Đặt: \(\left\{{}\begin{matrix}A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\\B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\end{matrix}\right.\)

Ta có:

\(B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{59}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{30}\right)\)

\(=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\)

\(\Rightarrow B=A\)

Vậy \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\) (Đpcm)

Ta có:

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+......+\dfrac{1}{59.60}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{59}-\dfrac{1}{60}\)

= \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{59}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)

- \(2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{60}\right)\) - \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)+ \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)

- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)

= \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)

Vậy\(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{59.60}\)

Do mọi số hạng của B đều lớn hơn 0 nên \(B>0\)

Lại có:

\(B=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\) (30 số hạng)

\(\Rightarrow B< \dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{31}+...+\dfrac{1}{31}\) (30 số hạng)

\(\Rightarrow B< \dfrac{1}{31}.30\)

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

Vậy \(0< B< 1\)

\(\Rightarrow B\) nằm giữa 2 số tự nhiên liên tiếp nên B không phải là số tự nhiên

Ta có:

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(\Rightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

Nhận xét:

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

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

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

\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{3}{5}\)

\(\Rightarrow S>\frac{3}{5}\left(1\right)\)

Lại có:

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

Nhận xét:

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{1}{3}\)

\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{1}{5}\)

\(\Rightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{4}{5}\)

\(\Rightarrow S< \frac{4}{5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)

\(\Rightarrow\frac{3}{5}< S< \frac{4}{5}\) (Đpcm)