fond lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } with a>0


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Question Number 35833 by abdo mathsup 649 cc last updated on 24/May/18
$fond lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } with a>0$
$f o n d {l i m}_{n \to + \infty} n^{a} {l n (1 + e^{- n}) - e^{- n}} w i t h a > 0$
Commented byprof Abdo imad last updated on 25/May/18
$we have for ∣x∣<1 ln^′ (1+x) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x) =Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) so ln(1+x) = x −(x^2 /2) +o(x^3 )(x→0)⇒ ln(1 +e^(−n) ) = e^(−n) − (e^(−2n) /2) +o(e^(−3n) )(n→+∞) n^a { ln(1+e^(−n) )−e^(−n) } ∼ −(1/2)n^a e^(−2n) (n→+∞) but lim_(n→+∞) n^a e^(−2n) =lim_(n→+∞) e^(aln(n)−2n) =lim_(n→+∞) e^(n{ a((ln(n))/n) −2}) = lim_(n→+∞) e^(−2n) =0 so for all a>0 lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } =0$
$w e h a v e f o r ∣ x ∣< 1 {l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}$ $\Rightarrow l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} x^{n}}{n}$ $s o l n (1 + x) = x - \frac{x^{2}}{2} + o (x^{3}) (x \to 0) \Rightarrow$ $l n (1 + e^{- n}) = e^{- n} - \frac{e^{- 2 n}}{2} + o (e^{- 3 n}) (n \to + \infty)$ $n^{a} {l n (1 + e^{- n}) - e^{- n}} \sim - \frac{1}{2} n^{a} e^{- 2 n} (n \to + \infty)$ $b u t {l i m}_{n \to + \infty} n^{a} e^{- 2 n} = {l i m}_{n \to + \infty} e^{a l n (n) - 2 n}$ $= {l i m}_{n \to + \infty} e^{n {a \frac{l n (n)}{n} - 2}} = {l i m}_{n \to + \infty} e^{- 2 n} = 0 s o$ $f o r a l l a > 0 {l i m}_{n \to + \infty} n^{a} {l n (1 + e^{- n}) - e^{- n}} = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18
	$when n→∞ then e^(−n) →0 so given limit (∞×0) form =((lim)/(n→∞))(({ln(1+e^(−n) )−e^(−n) })/(1/n^a )) ((0/0)) form using lhospitsl =((lim)/(n→∞))×(({(1/((1+e^(−n) )))×e^(−n) ×−1}−{e^(−n) ×(−1)})/(−a×n^(−a−1) )) =((lim)/(n→∞))×(({((−e^(−n) )/(1+e^(−n) ))}+e^(−n) )/(−a×n^(−a−1) )) =((lim)/(n→∞))×(({−e^(−n) +e^(−n) +e^(−2n) )/((1+e^(−n) )×(−a×n^(−a−1) ))) =(((−1)/a))×((lim)/(n→∞))(e^(−2n) /(1+e^(−n) ))×n^(a+1) =(((−1)/a))×((lim)/(n→∞))(1/(e^(2n) +e^n ))×n^(a+1) =(((−1)/a))×((lim)/(n→∞))(1/(e^n +1))×(n^(a+1) /e^n ) =((−/a))×((lim)/(n→∞))×((1/(e^(2n) +e^n ))/(1/n^(a+1) )) contd$
	$w h e n n \to \infty t h e n e^{- n} \to 0$ $s o g i v e n l i m i t (\infty \times 0) f o r m$ $= \frac{l i m}{n \to \infty} \frac{{l n (1 + e^{- n}) - e^{- n}}}{\frac{1}{n^{a}}} (\frac{0}{0}) f o r m$ $u s i n g l h o s p i t s l$ $= \frac{l i m}{n \to \infty} \times \frac{{\frac{1}{(1 + e^{- n})} \times e^{- n} \times - 1} - {e^{- n} \times (- 1)}}{- a \times n^{- a - 1}}$ $= \frac{l i m}{n \to \infty} \times \frac{{\frac{- e^{- n}}{1 + e^{- n}}} + e^{- n}}{- a \times n^{- a - 1}}$ $= \frac{l i m}{n \to \infty} \times \frac{{- e^{- n} + e^{- n} + e^{- 2 n}}{(1 + e^{- n}) \times (- a \times n^{- a - 1})}$ $= (\frac{- 1}{a}) \times \frac{l i m}{n \to \infty} \frac{e^{- 2 n}}{1 + e^{- n}} \times n^{a + 1}$ $= (\frac{- 1}{a}) \times \frac{l i m}{n \to \infty} \frac{1}{e^{2 n} + e^{n}} \times n^{a + 1}$ $= (\frac{- 1}{a}) \times \frac{l i m}{n \to \infty} \frac{1}{e^{n} + 1} \times \frac{n^{a + 1}}{e^{n}}$ $= (\frac{-}{a}) \times \frac{l i m}{n \to \infty} \times \frac{\frac{1}{e^{2 n} + e^{n}}}{\frac{1}{n^{a + 1}}}$ $c o n t d$
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