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Question Number 131167 by 676597498 last updated on 02/Feb/21

Answered by SEKRET last updated on 02/Feb/21

U_n =(1+(1/n^2 ))(1+(2/n^2 ))....(1+(n/n^2 ))  ln(U_n )= ln(1+(1/n^2 ))+ln(1+(2/(  n^2 )))+..+(1+(n/n^2 ))  ln(U_n )= Σ_(k=1) ^n ln(1+(k/n^2 ))  ln(U_n )= ln((((n^2 +1)_n )/n^(2n) ))    U_n = (((n^2 +1)_n )/n^(2n) )

Un=(1+1n2)(1+2n2)....(1+nn2)ln(Un)=ln(1+1n2)+ln(1+2n2)+..+(1+nn2)ln(Un)=nk=1ln(1+kn2)ln(Un)=ln((n2+1)nn2n)Un=(n2+1)nn2n

Commented by 676597498 last updated on 02/Feb/21

i dont get it

idontgetit

Answered by mathmax by abdo last updated on 02/Feb/21

u_n =Π_(k=1) ^n (1+(k/n^2 )) ⇒v_n =ln(u_n )=Σ_(k=1) ^n  ln(1+(k/n^2 ))  let prove first x−(x^2 /2)≤ln(1+x)≤x  let f(x)=x−ln(1+x) withx≥0  f^′ (x)=1−(1/(1+x))=(x/(1+x))≥0 ⇒f is increazinz but f(0)=0 ⇒f(x)≥0 ⇒  x≥ln(1+x) let g(x)=ln(1+x)−x+(x^2 /2) ⇒  g^′ (x)=(1/(1+x))−1+x =((1+x^2 −1)/(1+x))=(x^2 /(1+x))≥0  ⇒g is increazing  but  g(o)=0 ⇒g(x)≥0 ⇒x−(x^2 /2)≤ln(1+x)≤x ⇒  (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  ≤v_n ≤(1/n^2 )Σ_(k=1) ^n  k ⇒  (1/n^2 )×((n(n+1))/2)−(1/(2n^4 ))((n(n+1)(2n+1))/6)≤v_n ≤(1/n^2 )×((n(n+1))/2) ⇒  ((n^2  +n)/(2n^2 ))−(((n+1)(2n+1))/(12n^3 ))≤v_n ≤((n^2  +n)/(2n^2 )) ⇒  ((n+1)/(2n))−(((n+1)(2n+1))/(12n^3 ))≤v_n ≤((n+1)/(2n)) ⇒lim_(n→+∞) =(1/2) ⇒  lim_(n→+∞) ln(u_n )=(1/2) ⇒lim_(n→+∞) u_n =e^(1/2)  =(√e)

un=k=1n(1+kn2)vn=ln(un)=k=1nln(1+kn2)letprovefirstxx22ln(1+x)xletf(x)=xln(1+x)withx0f(x)=111+x=x1+x0fisincreazinzbutf(0)=0f(x)0xln(1+x)letg(x)=ln(1+x)x+x22g(x)=11+x1+x=1+x211+x=x21+x0gisincreazingbutg(o)=0g(x)0xx22ln(1+x)xkn2k22n4ln(1+kn2)kn21n2k=1nk12n4k=1nk2vn1n2k=1nk1n2×n(n+1)212n4n(n+1)(2n+1)6vn1n2×n(n+1)2n2+n2n2(n+1)(2n+1)12n3vnn2+n2n2n+12n(n+1)(2n+1)12n3vnn+12nlimn+=12limn+ln(un)=12limn+un=e12=e

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