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Question Number 88286 by Chi Mes Try last updated on 09/Apr/20

Commented by abdomathmax last updated on 09/Apr/20
$U_n ={(1+(1/n))^(n+1) −1−(1/n)}^(−n) ⇒U_n =e^(−nln{ (1+(1/n))^(n+1) −1−(1/n)} ) we have (1+(1/n))^(n+1) −1−(1/n) =e^((n+1)ln(1+(1/n))) −1−(1/n) and ln^′ (1+u)=1−u +o(u^2 ) ⇒ ln(1+u) =u−(u^2 /2) +o(u^3 ) ⇒ ln(1+(1/n))=(1/n)−(1/(2n^2 )) +o((1/n^3 )) ⇒ (n+1)ln(1+(1/n)) =1+(1/n)−((n+1)/(2n^2 )) +o((1/n^2 )) =1+(1/n)−(1/(2n))−(1/(2n^2 )) +o((1/n^2 )) ⇒ e^((n+1)ln(1+(1/n))) =e e^((1/(2n))−(1/(2n^2 ))+...) ∼e(1+(1/(2n))−(1/(2n^2 ))) ⇒ (1+(1/n))^(n+1) −1−(1/n) ∼e +(e/(2n))−(e/(2n^2 ))−1−(1/n) ∼e−1 +((e/2)−1)×(1/n)−(e/(2n^2 )) ⇒ U_n ∼e^(−n{ e−1 +((e−2)/(2n))−(e/(2n^2 ))}) =e^(−(e−1)n) e^(−((e−2)/2)+(e/(2n))) ∼e^((e−2)/2) ×e^(−(e−1)n) { 1+(e/(2n))} due to term e^(−(e−1)n) Σ U_n is convergent$
$U_{n} = {{(1 + \frac{1}{n})}^{n + 1} - 1 - \frac{1}{n}}^{- n}$ $\Rightarrow U_{n} = e^{- n l n {{(1 + \frac{1}{n})}^{n + 1} - 1 - \frac{1}{n}}} w e h a v e$ ${(1 + \frac{1}{n})}^{n + 1} - 1 - \frac{1}{n} = e^{(n + 1) l n (1 + \frac{1}{n})} - 1 - \frac{1}{n} a n d$ ${l n}^{^{'}} (1 + u) = 1 - u + o (u^{2}) \Rightarrow$ $l n (1 + u) = u - \frac{u^{2}}{2} + o (u^{3}) \Rightarrow$ $l n (1 + \frac{1}{n}) = \frac{1}{n} - \frac{1}{2 n^{2}} + o (\frac{1}{n^{3}}) \Rightarrow$ $(n + 1) l n (1 + \frac{1}{n}) = 1 + \frac{1}{n} - \frac{n + 1}{2 n^{2}} + o (\frac{1}{n^{2}})$ $= 1 + \frac{1}{n} - \frac{1}{2 n} - \frac{1}{2 n^{2}} + o (\frac{1}{n^{2}}) \Rightarrow$ $e^{(n + 1) l n (1 + \frac{1}{n})} = e e^{\frac{1}{2 n} - \frac{1}{2 n^{2}} + . . .} \sim e (1 + \frac{1}{2 n} - \frac{1}{2 n^{2}}) \Rightarrow$ ${(1 + \frac{1}{n})}^{n + 1} - 1 - \frac{1}{n} \sim e + \frac{e}{2 n} - \frac{e}{2 n^{2}} - 1 - \frac{1}{n}$ $\sim e - 1 + (\frac{e}{2} - 1) \times \frac{1}{n} - \frac{e}{2 n^{2}} \Rightarrow$ $U_{n} \sim e^{- n {e - 1 + \frac{e - 2}{2 n} - \frac{e}{2 n^{2}}}} = e^{- (e - 1) n} e^{- \frac{e - 2}{2} + \frac{e}{2 n}}$ $\sim e^{\frac{e - 2}{2}} \times e^{- (e - 1) n} {1 + \frac{e}{2 n}} d u e t o t e r m e^{- (e - 1) n}$ $Σ U_{n} i s c o n v e r g e n t$
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