Question Number 80614 by M±th+et£s last updated on 04/Feb/20
Commented by mathmax by abdo last updated on 04/Feb/20
![A_n =(1/n)Σ_(k=1) ^n ln(a+(k/n))⇒A_n is a Rieman sum and lim_(n→+∞) A_n =∫_0 ^1 ln(a+x)dx =_(a+x=t) ∫_a ^(1+a) ln(t)dt =[tln(t)−t]_a ^(1+a) =(1+a)ln(1+a)−(1+a)−(alna−a) =(1+a)ln(1+a)−1−a−alna+a =(1+a)ln(1+a)−alna −1](Q80634.png)
$${A}_{{n}} =\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left({a}+\frac{{k}}{{n}}\right)\Rightarrow{A}_{{n}} \:{is}\:{a}\:{Rieman}\:{sum} \\ $$$${and}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({a}+{x}\right){dx}\:=_{{a}+{x}={t}} \:\:\int_{{a}} ^{\mathrm{1}+{a}} {ln}\left({t}\right){dt} \\ $$$$=\left[{tln}\left({t}\right)−{t}\right]_{{a}} ^{\mathrm{1}+{a}} \:=\left(\mathrm{1}+{a}\right){ln}\left(\mathrm{1}+{a}\right)−\left(\mathrm{1}+{a}\right)−\left({alna}−{a}\right) \\ $$$$=\left(\mathrm{1}+{a}\right){ln}\left(\mathrm{1}+{a}\right)−\mathrm{1}−{a}−{alna}+{a} \\ $$$$=\left(\mathrm{1}+{a}\right){ln}\left(\mathrm{1}+{a}\right)−{alna}\:−\mathrm{1} \\ $$