1<1/n+1 +1/n+2 +......1/3n+1<2

sa se demonstreze folosinnd inductia

1<1/n+1 +1/n+2 +......1/3n+1<2

nu ma poate ajuta nimeni?

P(k): 1<1/(k+1)+1/(k+2)+⋯+1/(3k+1)<2
P(k+1): 1<1/(k+2)+1/(k+3)+⋯+1/(3k+4)<2
Prima etapa: Verificarea
pt k=1 avem
1<1/2+1/3+1/4<2 => 1<13/12<2 Adevarat
A doua etapa: Demonstratia
Daca P(k) adevarat implica P(k+1) adevarat
Ca sa fie mai simplu vom imparti problema in doua parti, mai precis :
1<1/(k+2)+1/(k+3)+⋯+1/(3k+4)
Si
1/(k+2)+1/(k+3)+⋯+1/(3k+4)<2
1<1/(k+1)+1/(k+2)+1/(k+3)+⋯+1/(3k+4)-1/(k+1)
1<1/(k+1)+1/(k+2)+1/(k+3)+⋯+1/(3k+1)+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
Stiind ca : 1<1/(k+1)+1/(k+2)+⋯+1/(3k+1) mai ramane sa demonstam ca 1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)>0
Aducand la numitor comun obtinem : 2/(3 ) 1/((3k+2)(3k+4)(k+1))>0 adevarat pentru ∀ k≥1 unde k∈N.
1<1/(k+2)+1/(k+3)+⋯+1/(3k+4) (I)
1/(k+1)+1/(k+2)+1/(k+3)+⋯+1/(3k+4)-1/(k+1)<2
1/(k+1)+1/(k+2)+1/(k+3)+⋯+1/(3k+1)+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)<2
Stiind ca : 1/(k+1)+1/(k+2)+⋯+1/(3k+1)<2 mai ramane sa demonstam ca 1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)>0
Uramrand rationamentul de mai sus obtinem: 2/(3 ) 1/((3k+2)(3k+4)(k+1))>0 adevarat pentru ∀ k≥1 unde k∈N.
1/(k+2)+1/(k+3)+⋯+1/(3k+4)<2 (II)
Din (I) si (II) obtinem:
1<1/(k+2)+1/(k+3)+⋯+1/(3k+4)<2 Adevarat