GENERALLY speaking ,usually in forming sequences in analysis there are two basic ways:

1)By using the axiom of Choice.

2)By using the inductive method

The following steps are a summary of the process in forming the sequence in concern,by using the axiom of Choice.

step 1) PROVE: For all nεΝ there exists an ,xε [a,b] and such that |f(x)|>n,by using the fact that f is unbounded over [a,b]

step 2) Form the sets ={x:xε[a,b],|f(x)|>n,nεN}

Step 3) By using step 1 prove : For all neΝ, .

step 4) Now use the axiom of Choice to form a sequence { } in [a,b] and such that |f( )|>n ,nεN.

Step 5) By the Bolzano Weiestrass theorem ,since the sequence { } is bounded in [a,b] ,there exists an accumulation point ,v in [a,b].

step 6)Prove:For all nεN ,THERE exists a , ,yε{ } and such that |y-v|<1/n,by using the fact that v is an accumulation point.

step 7)Form the sets, ={y:|y-v|<1/n,nεΝ, ,yε{ :nεN}}

Step 8) Prove: for all nεN by using step 6.

step 9) Use again the axiom of Choice to form the sequence { } in [a,b],such that |f( )|>n , and | -v|<1/n for all nεΝ

step 10) prove , .

ALTERNATIVELY,by using the inductive process we can say:

After we have proved that ,for all n belonging to natural Nos ,there exists an ,x belonging to [a,b] and such that |f(x)|>n.

For n= 1 ,we pick ,such that |f( )|>1.

For n=2 ,we pick ,such that |f( )|>2.

Suppose we have found:

all in [a,b] and such that |f( )|>n .

We select .We can thus proceed inductively to construct the sequence { },which is increasing and bounded from above by ,a.Hence there exists a,y in [a,b] and such, .

So we have proved the existence of a sequence { } in [a,b] such that |f( |>n and