Răspuns :
[tex]\displaystyle\it\\P(n)=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}<\frac{5n-2}{2n},~\forall n\in\mathbb{N},~n> 1.\\\underline{Etapa~verificarii:} \\n=1 \implies P(1): \frac{1}{1!}< \frac{5\cdot1-2}{2} =\frac{3}{2},~(A).\\\underline{Etapa~demonstratiei:}\\p.p~ca~P(k)~este~adevarata,~atunci~si~P(k+1)~este~adevarata.\\P(k):1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{k!}<\frac{5k-2}{2k}.\\P(k+1):1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{k!}+\frac{1}{(k+1)!}<\frac{5(k+1)-2}{2(k+1)}.[/tex]
[tex]\displaystyle\it\\\underbrace{1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{k!}}_{P(k)}+\frac{1}{(k+1)!}<\frac{5(k+1)-2}{2(k+1)} \Leftrightarrow\\\frac{5k-2}{2k}+\frac{1}{(k+1)!}<\frac{5(k+1)-2}{2(k+1)}~\bigg|-\frac{5k-2}{2k} \Leftrightarrow\\\frac{1}{(k+1)!}<\frac{1}{k^2+k} \Leftrightarrow (k+1)!>k^2+k \Leftrightarrow\\1\cdot2\cdot3\cdot...\cdot \underbrace{k\cdot (k+1)}_{k^2+k}>k^2+k,~evident,~pentru~n>1.\\deci~si~P(k+1)~este~adevarata,~in~concluzie~P(n)~este~adevarata \implies\\P(n+1)~este~adevarata.\\[/tex]
