study-the-convergence-of-n-1-ln-1-e-n-2-n-n-

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

s t u d y t h e c o n v e r g e n c e o f \sum_{n ⩾ 1} \frac{l n (1 + e^{- n^{2}})}{n^{n}}

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

l e t U_{n} = \frac{l n (1 + e^{- n^{2}})}{n^{n}} w e h a v e l n (1 + u) \sim u (u \in V (0)) \Rightarrow

l n (1 + e^{- n^{2}}) \sim e^{- n^{2}} \Rightarrow U_{n} \sim \frac{e^{- n^{2}}}{n^{n}}

f o r a l l ξ > 0 w e h a v e {l i m}_{n \to + \infty} n^{1 + ξ} U_{n} = {l i m}_{n \to + \infty} n^{1 + ξ} \frac{e^{- n^{2}}}{n^{n}}

= {l i m}_{n \to + \infty} \frac{e^{- n^{2}}}{n^{n - 1 - ξ}} = 0 \Rightarrow \exists A > 0 \forall n > n_{0} U_{n} < \frac{A}{n^{1 + ξ}}

t h e s e r i Σ \frac{A}{n^{1 + ξ}} i s c o n v e r g e n t \Rightarrow Σ U_{n} c o n v e r g e s .