sirurile fiind fractii , stabilim monotonia prin impartire si comparare cu 1
1 . a ( n +1 ) = [2· ( n +1 ) + 3 ] / [ 3·( n +1 ) -2 ] =
= [ 2n + 5 ] / [ 3n + 1 ]
impartire an /a(n +1 ) = ( 2n + 3 ) · [3n + 1] / [2n +5 ] · ( 3n -2 ) comparam cu 1
an / a( n +1) < 1
( 2n+ 3) · ( 3n +1 ) / ( 2n + 5) ·( 3n -2 ) < 1
( 2n+3) · ( 3n +1 ) / ( 2n +5)· ( 3n -2 ) - 1 <0
( 6n² +9n +2n +3 -6n² +4n -15n +6) / ( 2n+5) ·( 3n -2 ) < 0
n ∈ N 9 / ( 2n + 5) · ( 3n -2 ) < 0
↓ ↓
poz poz nu este negativ
presupunere falsa , deci an / a( n +1 ) > 1
an > a( n +1 ) sir monoton descrescator
marginirea - 3 / 2 ≤ an ≤ 2 / 3
2. an / a( n+1) = (3n +1 ) / ( n +1) / [ 3( n +1) + 1 ] / [ ( n +1 + 1 ]
= ( 3n +1 )·( n +2) / ( n+ 1) · ( 3n +4 ) < 1 preusupunere
( 3n + 1) ·( n +2) / ( n +1 ) ·( 3n +4 ) - 1 < 0
(3n² + n +6n + 2 - 3n² -3n - 4n - 4 ) / ( n +1 ) ·( 3n +4) < 0
- 2 / ( n +1 ) · ( 3n +4) < 0
↓ negativ < 0, adevarat
⇒ an / a( n +1 ) < 1
an < a( n +1 ) sir monoton crescator
marginirea 1 ≤ an ≤ 3 / 1
3. an = n² /( n² + 2)
a( n +1) = ( n +1) ² / [ ( n +1)² + 2] = ( n +1) ² / ( n² + 2n +3 )
calculam an - a( n +1) = [ n²·( n² +2n +3) - ( n +1) ²·( n² + 2) ] /(n²+2)·(n²+2n+3)
an - a( n +1) =( n⁴+2n³+3n²-n⁴-2n³-n²-2n²-4n-2) / ( n²+2) ·(n² + 2n +3) =
= - ( 4n +2) / ( n² +2) · ( n² + 2n +3 ) < 0
n∈N ↓ ↓ ↓ ↓
neg. poz. poz. poz.
⇒ an - a( n +1) < 0
an < a( n +1 ) monoton crescator
marginirea 0 ≤ an ≤ 1
