Limits of sequences
      Properties of convergent sequences
         Sandwich theorem (result) or squeeze rule
         Least upper bound (or supremum, abbrev. lub, sup) and greatest lower bound (infimum, abbrev. glb, inf)
      The definition of the real number e
Sandwich theorem (result) or squeeze rule
Suppose that {bn} is a sequence whose terms are bounded above and below (squeezed between) by sequences {cn} and {an} respectively, such that
an < bn < cn  for all n and if  an L  and  cn L,  then  bn L.
Least upper bound (or supremum, abbrev. lub, sup) and greatest lower bound (or infimum, abbrev. glb, inf)
Suppose {an} is a bounded increasing sequence of real numbers, then the least upper bound of the set { an :  n N } is the limit of {an}, so we write  L = sup an.
Proof:  Suppose  L = sup an  then, for given e > 0 there exists an integer n0 such that
  an0 > L - e   or   L - an0 < e.
Since {an} is increasing   an > an0 > L - e  for all  n > n0,  and since L is the upper bound of the sequence then,  L > an  for every n, therefore
| L - an | = L - an < L - an0 < e  for all  n > n0  that is,   an L.
Similarly, if {an} is a bounded decreasing sequence of real numbers, then the greatest lower bound of the set { an :  n N } is the limit of {an}, and we write  L = inf an.
The definition of the real number e
We use the monotonic sequence theorem to prove that the sequence defined by  
is increasing and its terms remain less than one fixed number as n tends to infinity, that is, the sequence converge to the number e.
Recall the binomial expansion theorem
Let expand an using the above theorem
Then, to prove an+1 > an, that is, that given is increasing sequence, we can expand an+1 the same way.
Observe that, while passing from n to n + 1, the an+1 expression gets one new positive term and, at the same time, all the differences in parentheses raised as every subtrahend decreased (i.e., denominators increased to n + 1). Therefore, since an+1 > an the sequence is increasing.
To show that all terms of the sequence are less than a fixed number M we will evaluate the quantity an on a suitable way.
If we omit second term (consisting 1/n) in every parenthesis, they increase, hence
If instead of the factors, 3, 4, . . . , n  in the denominators, starting from the second, we substitute 2, the denominators decreased so the fractions increased, thus
since applied is the formula for the sum of the finite geometric series whose ratio is 1/2 and the first term 1.
As the right side still depends on n, to get an expression independent of n, the one which holds to all terms, drop the term 1/2n in the numerator of the right side, what increases the numerator, such that the final result is
Therefore, as all terms of the sequence are less than M = 3 then, the sequence has a limit that is not greater than 3.
Let write few terms of the sequence to show how slowly it increases,
So, we finally write
Calculus contents B
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