Question

exercitiul 2 subpunctul c va rog ​

Answer 1

$x_{n}= 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-ln(n) \\x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}-ln(n+1)\\ x_{n+1}\leq x_{n}=>1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}-ln(n+1)\leq  1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-ln(n)<=>\frac{1}{n+1}-ln(n+1)\leq -ln(n) =>\frac{1}{n+1} \leq ln\frac{n+1}{n} ,\forall n \in N*=>x_{n}  \text{ sir descrescator}$

$\frac{x_{n+1} }{x_{n}}\leq1<=> \frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}-ln(n+1)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-ln(n) } \leq1<=>1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}-ln(n+1)\leq  1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-ln(n) =>\frac{1}{n+1}-ln(n+1) \leq -ln(n)=>\frac{1}{n+1} \leq ln(\frac{n+1}{n}) ,\foralln\in N*=>x_{n} \text{sir cu termeni pozitivi}$