Question

Calculati suma:
S=$\frac{1}{2!} +\frac{2}{3!} +...+\frac{99}{100!}$ si aratati ca S<1

Answer 1

$\it S=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\ ...\ +\dfrac{99}{100!} =\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+\ ...\ +\dfrac{100-1}{100!}=\\ \\ \\ =\Big(\dfrac{2}{2!}+\dfrac{3}{3!}+\dfrac{4}{4!}+\ ...\ +\dfrac{100}{100!}\Big)-\Big(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\ ...\ +\dfrac{1}{100!}\Big)=\\ \\ \\ =\Big(1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\ ...\ +\dfrac{1}{99!}\Big)-\Big(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\ ...\ +\dfrac{1}{100!}\Big)=\\ \\ \\ =1-\dfrac{1}{100!}<1$