CMR :
a) A = 1/2! + 1/3! + 1/4! + ... + 1/2020! < 1
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NC
1
22 tháng 4 2020
Viết sai rồi n!=1.2.3...n
Ta có \(\frac{1}{n!}=\frac{\left(n-1\right)!}{n!.\left(n-1\right)!}< \frac{\left(n-1\right).\left(n-1\right)!}{n!.\left(n-1\right)!}=\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
=> \(\frac{1}{2!}+\frac{1}{3!}+...\frac{1}{2020!}< \frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{2018!}-\frac{1}{2019!}+\frac{1}{2019!}-\frac{1}{2020!}\)
=> \(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2020!}< 1-\frac{1}{2020!}< 1\)(ĐPCM)
PT
1
4 tháng 4 2019
1/2! + 2/3! + 3/4! + ... + 2019/2020!
= (2-1) /2! + (3-1)/3! +(4-1)/4! +....+ (2019-1)/2019! + (2020-1)/2020!
= 2/2! -1/2! + 3/3! -1/3! + 4/4! -1/4! +..........+ 2019/2019! -1/2019! +2020/2020! -1/2020!
= 1/1! -1/2! + 1/2! -1/3! + 1/3! -1/4! +........+ 1/2018! -1/2019! +1/2019! -1/2020!
=1 -1/2020! <1
DT
0
XO
30 tháng 8 2020
Ta có \(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2020}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2019}{2020}=\frac{1.2.3...2019}{2.3.4...2020}=\frac{1}{2020}\)
Lại có : \(A=\left(1\frac{1}{2}\right).\left(1\frac{1}{3}\right).\left(1\frac{1}{4}\right)...\left(1\frac{1}{2020}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2021}{2020}=\frac{3.4.5...2021}{2.3.4...2020}=\frac{2021}{2}\)
Khi đó \(\frac{A}{B}=\frac{\frac{2021}{2}}{\frac{1}{2020}}=\frac{2021}{2}.2020=2041210\)
\(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2020!}\)
Ta có : \(\frac{1}{2!}=\frac{1}{1.2}\)
\(\frac{1}{3!}=\frac{1}{1.2.3}< \frac{1}{2.3}\)
\(\frac{1}{4!}=\frac{1}{1.2.3.4}< \frac{1}{3.4}\)
...
\(\frac{1}{2020!}=\frac{1}{1.2.3...2020}< \frac{1}{2019.2020}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A< 1-\frac{1}{2020}< 1\)
\(\Rightarrow\)A<1
Vậy A<1.
Mik làm giống bạn đs ó
P/s ; ko chắc