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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

An=1n∑k=1nln(a+kn)⇒AnisaRiemansumandlimn→+∞An=∫01ln(a+x)dx=a+x=t∫a1+aln(t)dt=[tln(t)−t]a1+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

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