limit of nondecreasing sequence
Proof. Let . Therefore the set has a finite supremum . We show that
Let an arbitrary positive number. According to the definition of supremum we have for all and on the other hand, there exists a member of the sequence that is . Then we have , and since the sequence is nondecreasing,
Thus the equation (1) and the whole theorem has been proven.
For the nonincreasing sequences there is the corresponding
Theorem. A monotonically nonincreasing sequence of real numbers with lower bound a number converges to a limit which is not less than .
Note. A good application of the latter theorem is in the proof that Euler’s constant exists.
|Title||limit of nondecreasing sequence|
|Date of creation||2013-03-22 17:40:31|
|Last modified on||2013-03-22 17:40:31|
|Last modified by||pahio (2872)|
|Synonym||nondecreasing sequence with upper bound|
|Synonym||limit of increasing sequence|