Question Number 114533 by 675480065 last updated on 19/Sep/20

Answered by abdomsup last updated on 19/Sep/20

u_n =Π_(k=1) ^n (1+(k/n^2 ))  a)let ϕ(x)=x−ln(1+x)  with x≥0 we have ϕ(0)=0  and ϕ^′ (x) =1−(1/(1+x)) =(x/(1+x))≥0  ⇒ϕ is increazing on[0,+∞[  ⇒x−ln(1+x)≥0 let  g(x)=ln(1+x)−x+(x^2 /2)  g(0)=0 and g^′ (x)=(1/(1+x))−1+x  =(1/(x+1)) +x−1 =((1+x^2 −1)/(x+1))=(x^2 /(x+1))≥0  ⇒g is inc .on[0,+∞[ ⇒  g≥0 ⇒ln)(1+x)≥x−(x^2 /2)  finally we get x−(x^2 /2)≤ln(1+x)  ≤x  we have ln(u_n )=Σ_(k=1) ^n ln(1+(k/n^2 ))  and (k/n^2 )−(k^2 /(2n^4 ))≤ln(1+(k/n^2 ))≤  (k/n^2 ) ⇒(1/n^2 )Σ_(k=1) ^n k−(1/(2n^4 ))Σ_(k=1) ^n  k^2   ≤Σ_(k=1) ^n  ln(1+(k/n^2 ))≤(1/n^2 )Σ_(k=1) ^n k ⇒  ((n(n+1))/(2n^2 ))−((n(n+1)(2n+1))/(12n^4 ))  ≤ln(u_n )≤((n(n+1))/(2n^2 ))  we passe to limit )n→+∞  (1/2) ≤lim ln(u_n )≤(1/2) ⇒  lim_(n→∞) ln(u_n )=(1/2)  ⇒lim_(n→∞) u_n =(√e)