Question

1/2!+2/3!+3/4!+5/6!+.....+2020/2021!<1 требуется доказать​

Answer 1

$\sum\limits_{n=0}^\infty \dfrac{x^n}{n!}=e^x\Rightarrow 1+x\sum\limits_{n=0}^\infty \dfrac{x^n}{(n+1)!}=e^x\Rightarrow \sum\limits_{n=0}^\infty \dfrac{x^n}{(n+1)!}=\dfrac{e^x-1}{x}\;\forall x\in R\backslash \{0\}$

$\sum\limits_{n=0}^\infty \dfrac{nx^{n-1}}{(n+1)!}=(\dfrac{e^x-1}{x})'_x \Rightarrow \sum\limits_{n=0}^\infty \dfrac{nx^{n-1}}{(n+1)!}=\dfrac{e^xx-(e^x-1)}{x^2}\Rightarrow \sum\limits_{n=1}^\infty \dfrac{n}{(n+1)!}=\sum\limits_{n=0}^\infty \dfrac{n*1^{n-1}}{(n+1)!}=\dfrac{e^1*1-(e^1-1)}{1^2}=1$

$\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2020}{2021!}<\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2020}{2021!}+\dfrac{2021}{2022!}\leq \sum\limits_{n=1}^\infty \dfrac{n}{(n+1)!}=1$

Ч.т.д.