límits no acotats y límits acotats

Definició:

[As][ s > 0 ==> [En_{0}][ n_{0}€N & [An][ n > n_{0} ==> a_{n} > s ] ] ]

[As][ s < 0 ==> [En_{0}][ n_{0}€N & [An][ n > n_{0} ==> a_{n} < s ] ] ]

Exemple:

[Ak][ k€N ==> n+k no és acotada-límit superiorment ].

Sigui s > 0 ==>

Es defineish n_{0} > s ==>

Sigui n > n_{0} ==>

n > s

Sigui k€N ==>

n+k >] n > s

Exemple:

[Ak][ k€N ==> (-n)+(-k) no és acotada-límit inferiorment ].

Sigui s < 0 ==>

Es defineish n_{0} > (-s) ==>

Sigui n > n_{0} ==>

n > (-s) 

(-n) < s

Sigui k€N ==>

(-n)+(-k) [< (-n) < s

Teorema:

Si [En_{0}][ n_{0}€N & [An][ n > n_{0} ==> a_{n} > n ] ] ==> ...

... a_{n} no és acotada-límit superiorment.

Sigui s > 0 ==>

Es defineish n_{0} = max{n_{1},n_{2}} & ...

... n_{1} > s & [An][ n > n_{2} ==> a_{n} > n ]

Sigui n > n_{0} ==>

a_{n} > n > s

Teorema:

Si [En_{0}][ n_{0}€N & [An][ n > n_{0} ==> a_{n} < (-n) ] ]  ==> ...

... a_{n} no és acotada-límit inferiorment.

Sigui s < 0 ==>

Es defineish n_{0} = max{n_{1},n_{2}} & ...

... n_{1} > (-s) & [An][ n > n_{2} ==> a_{n} < (-n) ]

Sigui n > n_{0} ==>

a_{n} < (-n) < s

Teorema:

Siguin a_{1},...,a_{n} > 0 ==> a_{1}+...+a_{n} no és acotada-límit superiorment.

Sigui s > 0 ==>

Es defineish M = min{a_{1},...,a_{n}} ==>

Es defienish n_{0} > (s/M) ==>

Sigui n > n_{0} ==>

a_{1}+...+a_{n} >] Mn > Mn_{0} > s

Teorema:

Siguin a_{1},...,a_{n} < 0 ==> a_{1}+...+a_{n} no és acotada-límit inferiorment.

Sigui s < 0 ==>

Es defineish M = max{a_{1},...,a_{n}} ==>

Es defienish n_{0} > (s/M) ==>

Sigui n > n_{0} ==>

a_{1}+...+a_{n} [< Mn < Mn_{0} < s

Teorema:

Si a_{n} & b_{n} no son acotades-límit superiorment ==> ...

... a_{n}+b_{n} no és acotada-límit superiorment.

Sigui s > 0 ==>

Sigui s_{1} = (s/2) & s_{2} = (s/2) ==>

Es defienish n_{0} = max{n_{1},n_{2}} & ...

... [An][ n > n_{1} ==> a_{n} > s_{1} ] & [An][ n > n_{2} ==> b_{n} > s_{2} ] ==>

Sigui n > n_{0} ==>

a_{n}+b_{n} > s_{1}+s_{2} = s

Teorema:

Si a_{n} & b_{n} no son acotades-límit inferiorment ==> ...

... a_{n}+b_{n} no és acotada-límit inferiorment.

Sigui s < 0 ==>

Sigui s_{1} = (s/2) & s_{2} = (s/2) ==>

Es defienish n_{0} = max{n_{1},n_{2}} & ...

... [An][ n > n_{1} ==> a_{n} < s_{1} ] & [An][ n > n_{2} ==> b_{n} < s_{2} ] ==>

Sigui n > n_{0} ==>

a_{n}+b_{n} < s_{1}+s_{2} = s

Problema-lema:

[Ak][ k€N ==> ( (n+k)^{2}/n ) no és acotada-límit superiorment ].

Sigui s > 0 ==>

Es defineish n_{0} > s ==>

Sigui n > n_{0} ==>

n > s

Sigui k€N ==>

( (n+k)^{2}/n ) = ( (n^{2}+2nk+k^{2})/n ) >] (n^{2}/n) > n > s

Problema-lema:

[Ak][ k€N ==> (-1)·( (n+k)^{2}/n ) no és acotada-límit inferiorment ].

Sigui s < 0 ==>

Es defineish n_{0} > (-s) ==>

Sigui n > n_{0} ==>

n > (-s)

(-n) < s

Sigui k€N ==>

(-1)·( (n+k)^{2}/n ) = (-1)·( (n^{2}+2nk+k^{2})/n ) [< (-1)·(n^{2}/n) < (-n) < s

Problema-lema:

[Ak][ k€N ==> ( n^{2}/(n+(-k)) ) no és acotada-límit superiorment ].

Sigui s > 0 ==>

Sigui k€N ==>

Es defineish n_{0} = max{n_{1},k} & n_{1} > s ==>

Sigui n > n_{0} ==>

n > s

( n^{2}/(n+(-k)) ) = ( ((n+(-k))^{2}+2(n+(-k))k+k^{2})/(n+(-k)) ) >] ...

... (n+(-k))+2k+( k^{2}/(n+(-k)) ) > n+k > n > s

Problema-lema:

[Ak][ k€N ==> (-1)·( n^{2}/(n+(-k)) ) no és acotada-límit inferiorment ].

Sigui s < 0 ==>

Sigui k€N ==>

Es defineish n_{0} = max{n_{1},k} & n_{1} > (-s) ==>

Sigui n > n_{0} ==>

n > (-s)

(-n) < s

(-1)·( n^{2}/(n+(-k)) ) = (-1)·( ((n+(-k))^{2}+2(n+(-k))k+k^{2})/(n+(-k)) ) = ...

... (-1)·( (n+(-k))+2k+( k^{2}/(n+(-k)) ) ) < (-n)+(-k) < (-n) < s

Definició:

[Es][ s > 0 ==> [An_{0}][ n_{0}€N ==> [En][ n > n_{0} & a_{n} [< s ] ] ]

[Es][ s < 0 ==> [An_{0}][ n_{0}€N ==> [En][ n > n_{0} & a_{n} >] s ] ] ]

Exemple:

[Ak][ k€N ==> ( 1/(n+k) ) està acotada superiorment ]

Es defienish s >] 1 ==>

Sigui n_{0}€N ==>

Es defineixh n = n_{0}+1 ==>

n > n_{0}

1 [< n_{0}+1 = n

( 1/(n+k) ) [< (1/n) [< 1 [< s

Exemple:

[Ak][ k€N ==> (-1)·( 1/(n+k) ) està acotada inferiorment ]

Es defienish s [< (-1) ==>

Sigui n_{0}€N ==>

Es defineixh n = n_{0}+1 ==>

n > n_{0}

1 [< n_{0}+1 = n

(-n) [< (-1)

(-1)·( 1/(n+k) ) >] (-1)·(1/n) >] (-1) >] s

Problema-lema:

[Ak][ k€N ==> ( 1/(n+(-k)) ) està acotada superiorment ]

Es defienish s >] 1 ==>

Sigui n_{0}€N ==>

Es defineixh n = (n_{0}+1)+k ==>

n > n_{0}

1 [< n_{0}+1

( 1/(n+(-k)) ) = ( 1/(n_{0}+1) ) [< 1 [< s

Problema-lema:

[Ak][ k€N ==> ( 1/(n+(-k)) ) està acotada superiorment ]

Es defienish s [< (-1) ==>

Sigui n_{0}€N ==>

Es defineixh n = (n_{0}+1)+k ==>

n > n_{0}

1 [< n_{0}+1

(-1)( n_{0}+1 ) [< (-1)

(-1)·( 1/(n+(-k)) ) = (-1)·( 1/(n_{0}+1) ) >] (-1) >] s

Teorema:

Si a_{n} & b_{n} son acotades-límit superiorment ==> ...

... a_{n}+b_{n} és acotada-límit superiorment.

Es defineish s > 0 & s_{1} = (s/2) & s_{2} = (s/2) ==>

Sigui n_{0}€N ==>

Siguin n_{1},n_{2}€N ==> 

Es defineish n = max{n_{0}+1,n_{1}+1,n_{2}+1} & ...

... [En][ n > n_{1} ==> a_{n} [< s_{1} ] & [En][ n > n_{2} ==> b_{n} [< s_{2} ] ==>

a_{n}+b_{n} [< s_{1}+s_{2} = s

Teorema:

Si a_{n} & b_{n} son acotades-límit inferiorment ==> ...

... a_{n}+b_{n} és acotada-límit inferiorment.

Es defineish s < 0 & s_{1} = (s/2) & s_{2} = (s/2) ==>

Sigui n_{0}€N ==>

Siguin n_{1},n_{2}€N ==> 

Es defineish n = max{n_{0}+1,n_{1}+1,n_{2}+1} & ...

... [En][ n > n_{1} ==> a_{n} >] s_{1} ] & [En][ n > n_{2} ==> b_{n} >] s_{2} ] ==>

a_{n}+b_{n} >] s_{1}+s_{2} = s