## Why I Don T Talk To Plan Your B2 First

Proof: Let all xn elements, at least starting with some number, satisfy to xnb inequality. Let's assume that and

It is possible to give one more definition to the meeting sequence, also: The sequence {xn} is called meeting if there is such number and that for any positive number of it is possible to specify number N such that at nN all xn elements of this sequence satisfy to an inequality:

Definition: The sequence {xn} is called meeting if there is such number and that the sequence {xn-and} is infinitesimal. Thus the number and is called as a limit of sequence {xn}.

Proof: Let an and b – limits of the meeting sequence {xn}. Then, using special representation for the xn elements of the meeting sequence {xn}, we will receive xn=a+n, xn=b+n, where n and n – elements of infinitesimal sequences {n} and {n}.

(owing to the theorem: Work of limited sequence on infinitesimal is infinitesimal sequence.) the sequence {a n+b n+n n} infinitesimal and therefore sequence {xnyn-b} too infinitesimal, so the sequence {xnyn} meets and has the limit ab number. The theorem is proved.

THEOREM: If elements of the meeting sequence {xn}, starting with some number, satisfy to a neraventstvo of xnb (xnb), and the limit and this sequence satisfies to an inequality of ab (ab).

If the general member of the row which is not meeting, dispersing in the true sense, aspires to zero, the partial sums of this row are located everywhere densely between their lower and top limits of lim inf and lim sup.

Subtracting these ratios, we will find n-n=b-a. As all elements of infinitesimal sequence {n-n} have the same constant b-a value, (according to the theorem: If all elements of infinitesimal sequence {n} are equal to the same number with, with = to b-a=0, i.e. b=a. The theorem is proved.

follows from this inequality that at nN |yn inequality | is carried out>. Therefore at nN we have. Therefore, starting with this number N, we can consider sequence, and this sequence is limited. The lemma is proved.

The sequence meets and has the limit zero. After all what there was > 0, on Archimedes's property of real numbers there is such natural number of n that n>. Therefore for all nn, and it means that.

THEOREM: Let {xn} and {zn} - the meeting sequences having the general limit and. Let, besides, starting with some number, elements of sequence {yn} satisfy to xnynzn inequalities. Then the sequence {yn} meets and has a limit and.

Proof: it is enough to prove that {yn-a} is infinitesimal. Let's designate through N’ number starting with which, the inequalities specified in a theorem condition are carried out. Then, starting with the same number, will be also inequalities of xn-á yn-á zn-á will be executed. From this it follows that at nN’ elements of sequence {yn-a} satisfy to an inequality

where n-element of infinitesimal sequence. As the infinitesimal sequence {n} is limited (according to the theorem: The infinitesimal sequence is limited.), there will be such number A that for all numbers n fairly |n|A inequality. Therefore | xn | |a | + A for all numbers n, as means limitation of sequence {xn}. The theorem is proved.