Inductie matematica
a)
pt n=1 verificam: 1/2+1/3+1/4 = 6+4+3 / 12 =13/12>1 ok
pp enuntul adevarat
dem pt n=k+1
1/(k+2) + 1/(k+3) + ... + 1/(3k+4) = 1/(k+1) +1/(k+2) +...+1/(3k+1) - 1/(k+1) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) >(cf ip inductie) 1 - 1/(k+1) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) >
acum ne folosim de inegalitatea 1/(k+1) < 1/(3k+2) si amplificand cu -1 avem
-1/(k+1) > -1/(3k+2) si revenind mai sus, avem:
1 - 1/(k+1) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) > 1 - 1/(3k+2) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) > 1 + 1/(3k+3) + 1/(3k+4) > 1 deci presupunerea facuta pt n=k este adevarata si pt n=k+1, ∀n,k∈N, n,k≥1.
b)
pt n=1: 1≤1, se verifica
pp adev pt n=k
v.d. pt n=k+1
1+1/3+1/7 +...+1/(2^k -1) + 1/(2^(k+1) - 1) ≤(cf ip ind) k+ p ( si cum p≤1 )⇒≤ k+1
c)
pt n=1: 1/2>13/24≅0,54 FALS, deci enuntul de la c este FALS.