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Question Number 164591 by SANOGO last updated on 19/Jan/22

pour quelle valeur α la serie converge  Σ_(n=2) (ln(n)+αln(n−(1/n))

$${pour}\:{quelle}\:{valeur}\:\alpha\:{la}\:{serie}\:{converge} \\ $$$$\underset{{n}=\mathrm{2}} {\sum}\left({ln}\left({n}\right)+\alpha{ln}\left({n}−\frac{\mathrm{1}}{{n}}\right)\right. \\ $$

Answered by mindispower last updated on 19/Jan/22

lim_(n→∞) ln(n)+aln(n−(1/(n)))  =lim_(n→∞) (1+a)ln(n)+aln(1−(1/n^2 ))  =lim_(n→∞) (1+a)ln(n)= { ((0 if a=−1)),((+_− ∞)) :}a≠0  if a=−1  ln(n)−ln(n−(1/n))=ln((n^2 /(n^2 −1)))=ln(1+(1/(n^2 −1)))  =(1/(n^2 −1))+o((1/n^2 )),n→∞  Σ(1/(n^2 −1)) cv ⇒Σln(n)−ln(n−(1/n)) cv  Σ_(n≥2) ln((n^2 /(n^2 −1)))=ln(Π_(N≥n≥2) (n^2 /((n−1)(n+1))))  =ln((N/1).(2/(N+1)))  lim_(N→∞) ln(((2N)/(N+1)))=ln(2)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{ln}\left({n}\right)+{aln}\left({n}−\frac{\mathrm{1}}{\left.{n}\right)}\right. \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+{a}\right){ln}\left({n}\right)+{aln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+{a}\right){ln}\left({n}\right)=\begin{cases}{\mathrm{0}\:{if}\:{a}=−\mathrm{1}}\\{\underset{−} {+}\infty}\end{cases}{a}\neq\mathrm{0} \\ $$$${if}\:{a}=−\mathrm{1} \\ $$$${ln}\left({n}\right)−{ln}\left({n}−\frac{\mathrm{1}}{{n}}\right)={ln}\left(\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}\right)={ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right),{n}\rightarrow\infty \\ $$$$\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\:{cv}\:\Rightarrow\Sigma{ln}\left({n}\right)−{ln}\left({n}−\frac{\mathrm{1}}{{n}}\right)\:{cv} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}{ln}\left(\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}\right)={ln}\left(\underset{{N}\geqslant{n}\geqslant\mathrm{2}} {\prod}\frac{{n}^{\mathrm{2}} }{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}\right) \\ $$$$={ln}\left(\frac{{N}}{\mathrm{1}}.\frac{\mathrm{2}}{{N}+\mathrm{1}}\right) \\ $$$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}{ln}\left(\frac{\mathrm{2}{N}}{{N}+\mathrm{1}}\right)={ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

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