Question

\(CMR:\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1\)

Answer 1

\(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)

=\(\frac{2-1}{2!}+\frac{3-2}{3!}+...+\frac{100-99}{100!}\)

\(=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{2}{3!}+...+\frac{100}{100!}-\frac{99}{100!}\)

\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{100!}-\frac{1}{99!}\)

\(=1-\frac{99}{100!}< 1\)

\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1\left(đpcm\right)\)

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