chứng tỏ rằng S = \(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....+\frac{1}{2001!}\)< 3
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HT
Chứng tỏ rằng :
S = \(\frac{1}{1!}\)+ \(\frac{1}{2!}\)+ \(\frac{1}{3!}\)+ ...+ \(\frac{1}{2001!}\)< 3
1
7 tháng 5 2017
Ta có:
1/1! = 1
1/2! = 1/1.2
1/3! = 1/2.3
1/4! < 1/3.4
1/5! < 1/4.5
.........
1/2001! < 1/2000.2001
==> S < 1 + 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2000.2001
S < 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2000 - 1/2001
S < 1 + 1 - 1/2001
S < 2 - 1/2001 < 2 < 3
==> S < 3
3 tháng 3 2017
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
7 tháng 8 2018
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2S=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{2018}}\right)\)
\(S=1-\frac{1}{2^{2018}}< 1\)
7 tháng 8 2018
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2018}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2S-S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2018}}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2018}}\)
\(S=1-\frac{1}{2^{2018}}\)
\(Mà
1-\frac{1}{2^{2018}}< 1\)
\(\Rightarrow S< 1\)
TN
1
10 tháng 4 2017
VÌ \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...........;\frac{1}{99^2}=\frac{1}{99\cdot99}< \frac{1}{99\cdot100}\)
\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}< 1\)\(\Rightarrow S< 1\)
VÌ \(\frac{1}{2\cdot3}< \frac{1}{2\cdot2};.....;\frac{1}{98\cdot99}< \frac{1}{99\cdot99}\)
\(\Rightarrow\)\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{98\cdot99}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}< S\)
\(\Rightarrow\frac{49}{100}< S< 1\)
\(K\)\(mk\)\(nha\)
9 tháng 4 2019
Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)
\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)
\(\Rightarrow S< 2\) ( ĐPCM )
19 tháng 5 2021
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)