Question

sa se demonstreze inegalitatea
1/(n+1)+1/(n+2)+......+1/2n>13/24 unde n€N*, n>2

Answer 1

 
Text corect;
unde n€N*, n≥2


verificam pt n=2
1/3+1/4=7/12=14/24>13/24
Presupunem adevarat pt n, n≥2
notam cu Sn  expresia pt n si cu Sn+1 expresia pt n+1

verificam pt n->n+1
 S n+1= 1/(n+2)+1/(n+3)+......1/2n+1/(2n+1)+1/(2n+2)=
Sn -1/(n+1)+1/2n+1/(2n+1)+1/(2n+2),
 atunci pt Sn+1 avem de demonstrat ca Sn+1>13/24

Sn+1=Sn+1/2n+1/(2n+1)+1/(2n+2)-1/(n+1)=Sn+a(n), unde prin a(n) am notata numarul
1/2n+1/(2n+1)+1/(2n+2)-1/(n+1)


 este sufiecient sa demonstram ca a(n)>0
 adica
1/2n+1/(2n+1)+1/(2n+2)-1/(n+1)>0
dar
1/2n+1/(2n+1)+1/(2n+2)>1/(2n+2)>1/(2n+2)+1/(2n+2)=3/(2(n+1))
deci
a(n)=1/2n+1/(2n+1)+1/(2n+2)-1/(n+1)>3/(2(n+1)) -1/(n+1)=1/(2(n+1))>0
deci a(n) >0
atunci
Sn+1=Sn+a(n)
cum  Sn>13/24 si a(n) >0⇒Sn+1>13/24
asadar inegalitatea este demonstrata prin inductie matematica completa