biết rằng n!=1.2.3....n.
chứng minh rằng 5/3<1/1!+1/2!+...+1/2016!<2
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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 )
Do a < b < c < d < m < n
=> a + c + m < b + d + n
=> 2.(a + c + m) < a + b + c + d + m + n
=> \(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\) (đpcm)
ta có:1/2!<1
2/3!<1
......
......
2015/2016!<1
=>A=1/2!+2/3!+3/4!+......+2015/2016! luôn luôn <1
m:n = 1+1/2+1/3+...+1/2016
m=(1+1/2+1/3+...+1/2016) . n
m=(1+1/2016) +(1/2+1/2015) +(1/3+1/2014) +...+(1/1008+1/1009). n
m=2017/2016 +2017/(2x2015) +2017/(3x2014)+...+(2017/1008x1009). n
m=2017x(1/2016+1/(2x2015)+1/(3x2014)+...+1/(1008x1009) . n
Vậy m chia hết cho2017