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\(M=0.5-\dfrac{2}{3!}-\dfrac{3}{4!}-...-\dfrac{2013}{2014!}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{2014}\)
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 )
Ta có: \(\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
Áp dụng vào M ta được:
\(M=\frac{1}{2!}-\frac{2}{3!}-\frac{3}{4!}-\frac{4}{5!}-...-\frac{2013}{2014!}\)
\(=\frac{1}{2!}-\left(\frac{1}{2!}-\frac{1}{3!}\right)-\left(\frac{1}{3!}-\frac{1}{4!}\right)-...-\left(\frac{1}{2013!}-\frac{1}{2014!}\right)\)
\(=\frac{1}{2!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{3!}+\frac{1}{4!}-...-\frac{1}{2013!}+\frac{1}{2014!}=\frac{1}{2014!}\)