Question

1/1*4+1/2*5+1/3*6+....+1/n(n+3)
Mărginire sir.Va rooog.

Answer 1

1/1*4>0
1/2*5>0
...........
1/n(n+3) >0
an>0

1/1*4=(1/3)(1-1/4)
1/2*5=(1/3)(1/2-1/5)
1/(3*6)=(1/3)(1/3-1/6)
nu merge ca am serie armonica in stanga ce e divergenta; termeniinu se mai reduc

deci facem un artificiu, majoram sirul
1/1*4<1/1*2=1-1/2
1/2*5<1/2*3=1/2-1/3
1/3*6<1/3*4=1/3-1/4
.........................................
1/(n-1)(n+2)<1/(n-1)n=1/n-1-1/n
1/n(n+3)<1/n(n+1)=1/n-1/(n+1)
adunam pe coloana

an<1-1/(n+1)<1
deci
0<an<1., asadar an marginit, C.C.T .D.
 cum e si crescator, are si limita
dar habar nu am care este, bine ca nu ne-o cere