Cho S= 1/11+ 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/19 + 1/20
So sánh S với 1/2
Mn giúp mình với
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Ta có:\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\frac{1}{13}>\frac{1}{20};....;\frac{1}{19}>\frac{1}{20}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(Có 10 phân số \(\frac{1}{20}\))
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{10}{20}\)\(\Leftrightarrow S>\frac{10}{20}\)
Mà \(\frac{10}{20}=\frac{1}{2}\)nên
\(\Rightarrow S>\frac{1}{2}\)
Ta có các phân số 1/11 ; 1/12 ; 1/13 ; 1/14 ; 1/15 ; 1/16 ; 1/17 ; 1/18 ; 1/19 đều lớn hơn 1/20
Do đó : 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/19 + 1/20 > 1/20 + 1/20 + ;...+ 1/20 ( có 10 phân số 1/20 )
1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1 /16 + 1/17 + 1/18 + 1/19 + 1/20 > 10/20
1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1 /16 + 1/17 + 1/18 + 1/19 + 1/20 > 1/2
Vậy : S > 1/2
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+...+\frac{1}{20}\)(10 số \(\frac{1}{20}\))
=\(\frac{1}{20}.10=\frac{1}{2}\)
vậy S>1/2
\(S=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(10 số hạng)
\(=10.\frac{1}{20}=\frac{1}{2}\).
Vậy \(S>\frac{1}{2}\).
ta thấy: 1/11;1/12;1/13;...;1/19;1/20 đều >1/20
=>1/11+1/12+...1/19+1/20>1/20+1/20...+1/20
1/11+1/12+...1/19+1/20>10/20
1/11+1/12+...1/19+1/20>1/2 vậy S>1/2
Ta thấy mỗi số hạng của tổng đều bé hơn 1/10
=>S<\(\frac{1}{10}.10=1\)
=>S<1
S = 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/19 + 1/20
S < 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10
S < 10 × 1/10
S < 1
có \(\frac{1}{20}\) bé nhất suy ra
"có 10 số hạng "\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+......+\frac{1}{20}>\frac{1}{20}.10\)
\(VT>\frac{10}{20}=\frac{1}{2}\)
vì 1/11>1/20
1/12>1/20...
1/13>1/20
nên 1/11+1/12+,,,,+1/20>1/20+1/20+,,,,+1/20=10/20=1/2(rút gọn
10 số 1/20
vậy S>1/2
\(\frac{1}{11}\)> \(\frac{1}{20}\)
\(\frac{1}{12}\)> \(\frac{1}{20}\)
.
.
.
\(\frac{1}{19}\)>\(\frac{1}{20}\)
\(\frac{1}{20}\)= \(\frac{1}{20}\)
=> S = 1/11+1/12+...+1/20>1/20+1/20+1/20+1/20+1/20+1/20+1/20+1/20+1/20+1/20=10*1/20=1/2 (đpcm)
Ta có:
\(\frac{1}{11}>\frac{1}{20}\)
\(\frac{1}{12}>\frac{1}{20}\)
.............
\(\frac{1}{20}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\) ( 10 phân số \(\frac{1}{20}\))
\(\Leftrightarrow\frac{10.1}{20}=\frac{10}{20}=\frac{1}{2}\)
Vì \(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{2}\). Mà \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{2}\)
-Ta có: \(\dfrac{1}{2}=\dfrac{10}{20}=\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (có 10 số \(\dfrac{1}{20}\)).
Mà \(\dfrac{1}{20}< \dfrac{1}{19};\dfrac{1}{20}< \dfrac{1}{18};...;\dfrac{1}{20}< \dfrac{1}{11}\)
\(\Rightarrow\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}< \dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{11}\)
\(\Rightarrow\dfrac{1}{2}< S\)