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Question Number 139530 by mnjuly1970 last updated on 28/Apr/21
.......advanced calculus...... prove that:: lim_(n→∞) {(((−1)^(n+1) n^(n+1) )/(n!)) (d^( n) /dx^n )(((ln(x))/x))∣_(x=n) }=γ γ : euler −mascheroni constant
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…….{advanced}\:\:{calculus}…… \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:{lim}_{{n}\rightarrow\infty} \left\{\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}^{{n}+\mathrm{1}} }{{n}!}\:\frac{{d}^{\:{n}} }{{dx}^{{n}} }\left(\frac{{ln}\left({x}\right)}{{x}}\right)\mid_{{x}={n}} \right\}=\gamma \\ $$$$\:\gamma\::\:\:\:{euler}\:−{mascheroni}\:{constant} \\ $$$$ \\ $$
Answered by mindispower last updated on 28/Apr/21
(d^n /dx^n )((ln(x))/x)=Σ_(k=0) ^n C_n ^k (ln(x))^k .((1/x))^(n−k) .Libneiz formula ln(x)^((k)) =ln(x),k=0 =((1/x))^((k−1)) ,k≥1 =(((−1)^(k−1) .(k−1)!)/x^(k−1) ),k≥1 ((1/x))^(n−k) =(((−1)^(n−k) (n−k)!)/x^(n−k+1) ) =(Σ_(k=1) ^n C_n ^k (((−1)^(n−1) (k−1)!.(n−k)!)/x^(n+1) )+ln(x).(((−1)^n n!)/x^(n+1) ))∣_(x=n) .(((−1)^(n+1) n^(n+1) )/(n!)) =Σ_(k≥1) (((−1)^(n−1) )/n^(n+1) ).(((k−1)!.(n−k)!)/n^(n+1) ).((n!)/(k!.(n−k)!)).(((−1)^(n+1) )/(n!))n^(n+1) +ln(n).(((−1)^n )/n^(n+1) )n!.(((−1)^(n+1) n^(n+1) )/(n!)) =Σ_(k=1) ^n (1/k)−ln(n) lim_(n→∞) Σ_(k=1) ^n (1/k)−ln(n)=γ By definition
$$\frac{{d}^{{n}} }{{dx}^{{n}} }\frac{{ln}\left({x}\right)}{{x}}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left({ln}\left({x}\right)\right)^{{k}} .\left(\frac{\mathrm{1}}{{x}}\right)^{{n}−{k}} .{Libneiz}\:{formula} \\ $$$${ln}\left({x}\right)^{\left({k}\right)} ={ln}\left({x}\right),{k}=\mathrm{0} \\ $$$$=\left(\frac{\mathrm{1}}{{x}}\right)^{\left({k}−\mathrm{1}\right)} ,{k}\geqslant\mathrm{1} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} .\left({k}−\mathrm{1}\right)!}{{x}^{{k}−\mathrm{1}} },{k}\geqslant\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{{n}−{k}} =\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{{x}^{{n}−{k}+\mathrm{1}} } \\ $$$$=\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({k}−\mathrm{1}\right)!.\left({n}−{k}\right)!}{{x}^{{n}+\mathrm{1}} }+{ln}\left({x}\right).\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{x}^{{n}+\mathrm{1}} }\right)\mid_{{x}={n}} .\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}^{{n}+\mathrm{1}} }{{n}!} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{{n}+\mathrm{1}} }.\frac{\left({k}−\mathrm{1}\right)!.\left({n}−{k}\right)!}{{n}^{{n}+\mathrm{1}} }.\frac{{n}!}{{k}!.\left({n}−{k}\right)!}.\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}!}{n}^{{n}+\mathrm{1}} \\ $$$$+{ln}\left({n}\right).\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{n}+\mathrm{1}} }{n}!.\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}^{{n}+\mathrm{1}} }{{n}!} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}−{ln}\left({n}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}−{ln}\left({n}\right)=\gamma\:\:\:{By}\:{definition} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 28/Apr/21
thanks alot sir power ... mercey....
$${thanks}\:{alot}\:{sir}\:{power}\:… \\ $$$${mercey}…. \\ $$
Commented by mindispower last updated on 29/Apr/21
pleasur
$${pleasur} \\ $$

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