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Question Number 62204 by maxmathsup by imad last updated on 17/Jun/19

$$studytheconvergenceof\sum _{n⩾1}\frac{ln(1+e^{-n^2})}{n^n}$$

Commented by maxmathsup by imad last updated on 19/Jun/19

$$let{U}_n=\frac{ln(1+e^{-n^2})}{n^n}wehaveln(1+u)\sim u(u\in V\left(0\right))\Rightarrow$$$$ln(1+e^{-n^2})\sim e^{-n^2}\Rightarrow U_n\sim \frac{e^{-n^2}}{n^n}$$$$forall\xi >0wehave{lim}_{n\to +\mathrm{\infty }}{n}^{1+\xi }{U}_n={lim}_{n\to +\mathrm{\infty }}{n}^{1+\xi }\frac{e^{-n^2}}{n^n}$$$$={lim}_{n\to +\mathrm{\infty }}\frac{e^{-n^2}}{n^{n-1-\xi }}=0\Rightarrow \mathrm{\exists }A>0\mathrm{\forall }n>n_0{U}_n<\frac{A}{n^{1+\xi }}$$$$theseri\mathrm{\Sigma }\frac{A}{n^{1+\xi }}isconvergent\Rightarrow \mathrm{\Sigma }{U}_{n}converges.$$

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