So sánh \(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\) với \(\frac{7}{4}\) (kí hiệu \(n!=1.2.3...n\))
K
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14 tháng 11 2018
\(P=\frac{3}{1!\left(1+2\right)+3!}+\frac{4}{2!\left(1+3\right)+4!}+...+\frac{2017}{2015!\left(1+2016\right)+2017!}\)
\(P=\frac{3}{3\left(1!+2!\right)}+\frac{4}{4\left(2!+3!\right)}+...+\frac{2017}{2017\left(2015!+2016!\right)}\)
\(P=\frac{1}{1!+2!}+\frac{1}{2!+3!}+...+\frac{1}{2015!+2016!}\)
Ta có \(a!>\sqrt{a}\)\(\left(a\inℕ;a>1\right)\) do đó :
\(P>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\)
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2016}\)
\(-\sqrt{2015}=\sqrt{2016}-1=\frac{1}{2}+\left(\sqrt{2016}-\frac{3}{2}\right)=\frac{1}{2}+\left(\sqrt{2016}-\sqrt{\frac{9}{4}}\right)>\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Chúc bạn học tốt ~
PS : tự nghĩ bừa thui nhé :))
12 tháng 3 2020
Ta có: \(\frac{1}{5!}=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}< \frac{1}{3\cdot4\cdot5}\)
\(\frac{1}{6!}< \frac{1}{1\cdot2\cdot3\cdot4\cdot5\cdot6}< \frac{1}{4\cdot5\cdot6}\)
..............
\(\frac{1}{2019!}=\frac{1}{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2019}< \frac{1}{2017\cdot2018\cdot209}\)
Do đó
\(C< 1+\frac{1}{2}+\frac{1}{2\cdot3\cdot4}+\frac{1}{4\cdot5\cdot6}+....+\frac{1}{2017\cdot2018\cdot2019}\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+.....+\frac{2019-2017}{2017\cdot2018\cdot2019}\right)\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2018\cdot2019}\right)< \frac{3}{2}+\frac{1}{2}\cdot\frac{1}{1\cdot2}\)
\(\Rightarrow C< \frac{7}{4}\)
Nguồn: Nock Nock
K
22 tháng 2 2020
\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
\(=\frac{1}{1}+\frac{1}{1.2}+\frac{1}{1.2.3}+...+\frac{1}{1.2.3...2019}\)
\(=\frac{1}{1}+\frac{1}{1}.\frac{1}{2}+\frac{1}{1}.\frac{1}{2}.\frac{1}{3}+...+\left(\frac{1}{1}.\frac{1}{2}.\frac{1}{3}...\frac{1}{2018}.\frac{1}{2019}\right)\)
\(=\left(1.1.1....1.1\right)+\left(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}\right)+\left(\frac{1}{3}.\frac{1}{3}.\frac{1}{3}...\frac{1}{3}.\frac{1}{3}\right)+...+\left(\frac{1}{2018}.\frac{1}{2018}\right)+\frac{1}{2019}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}+\frac{1}{2019}\)
Nhận xét rằng:
\(1< \frac{7}{8076};2< \frac{7}{8076};3< \frac{7}{8076};...;\frac{1}{1154}>\frac{7}{8076};\frac{1}{1155}>\frac{7}{8076};...;\frac{1}{2018}>\frac{7}{8076};\frac{1}{2019}>\frac{7}{8076}\)
Do đó:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}+\frac{1}{2019}>\frac{7}{8076}+\frac{7}{8076}+...+\frac{7}{8076}\)
Vì tổng C có 2019 số hạng, suy ra \(C>2019.\frac{7}{8076}=\frac{7}{4}\)
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 )
12 tháng 3 2020
Tham khảo nhé
Câu hỏi của Assassin_07 - Toán lớp 7 - Học toán với OnlineMath
LT
3 tháng 2 2020
GIẢ SỬ \(\frac{A}{B}=\frac{C}{D}\)
ĐẶT\(\frac{A}{B}=\frac{C}{D}=T\)=>A = BT , C = DT
TA CÓ\(\frac{\left(A^2+B^2\right)}{\left(C^2+D^2\right)}=\frac{\left(\left(B\cdot T\right)^2+B^2\right)}{\left(\left(D\cdot T\right)^2+D^2\right)}=\frac{\left(B^2\cdot\left(T^2+1\right)\right)}{\left(D^2\cdot\left(T^2+1\right)\right)}=\frac{B^2}{D^2}=\left(\frac{B}{D}\right)^2\left(1\right)\)
LẠI CÓ\(\frac{\left(A\cdot B\right)}{\left(C\cdot D\right)}=\frac{\left(B\cdot T\cdot B\right)}{\left(D\cdot T\cdot D\right)}=\frac{B^2}{D^2}=\left(\frac{B}{D}\right)^2\left(2\right)\)
TỪ (1) VÀ (2) \(\Rightarrow\frac{\left(A^2+B^2\right)}{\left(C^2+D^2\right)}=\frac{\left(A\cdot B\right)}{\left(C\cdot D\right)}\)( THÕA ĐỀ )
=> ĐIỀU GIẢ SỬ ĐÚNG => DPCM
XO
12 tháng 8 2019
B = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}...+\frac{1}{1+2+3+...+2019}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2019\times1010}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2019\times2020}\right)\)
= \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2019\times2020}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=2\times\frac{1009}{2020}\)
\(=\frac{1009}{1010}< \frac{1010}{1010}=1\)
\(\Rightarrow B< 1\)
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