Zad. 2.Rozwiąż nierówność, w której lewa strona jest sumą nieskończonego ciągu geometrycznego:
Mamy
a1 = 1/ ( x + 1)
oraz
q = 1/(x + 1) , gdzie I q I < 1
zatem
S = a1/( 1 - q) = [ 1 /(x +1)]/ [ 1 - 1/(x + 1)]
S = [ 1 /(x + 1)]/[ (x+1)/(x +1) - 1/(x + 1)]
S = [ 1 / ( x+1)] /[ x /( x + 1)] =1/x
Mamy więc nierówność:
1/x < 3
x > 1/3
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Mamy
a1 = 1/ ( x + 1)
oraz
q = 1/(x + 1) , gdzie I q I < 1
zatem
S = a1/( 1 - q) = [ 1 /(x +1)]/ [ 1 - 1/(x + 1)]
S = [ 1 /(x + 1)]/[ (x+1)/(x +1) - 1/(x + 1)]
S = [ 1 / ( x+1)] /[ x /( x + 1)] =1/x
Mamy więc nierówność:
1/x < 3
x > 1/3
=================