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Pentru orice intreg n demonstrati inegalitatea :
\( \frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{n+2}{n!+(n+1)!+(n+2)!}\le \frac{1}{2} \) .

Studiem termenul general:\( \frac{n+2}{n!+(n+1)!+(n+2)!}=\frac{n+2}{n!(n+2)+(n+2)!}=\frac{1}{n!+(n+1)!}=\frac{1}{n!(n+2)}=\frac{n+1}{(n+2)!}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!} \)
Acum \( \sum{\frac{n+2}{n!+(n+1)!+(n+2)!}}=\frac{1}{2}-\frac{1}{(n+2)!}\le\frac{1}{2} \)

Primul termen este \( \frac{3}{1!+2!+3!}=\frac{3}{1+2+6}=\frac{1}{3} \).

Da , primul termen este \( \frac{1}{3} \) . Nu inteleg ce vreti sa spuneti .

Ce, am gresit? Mie imi pare corect!

Pai , da , la aceeiasi rezolvare m-am gandit si eu . Nu inteleg insa ce rost are observatia domnului mihai++ .