# limit of nondecreasing sequence

Proof.  Let  $a_{1}\leqq a_{2}\leqq\ldots\leqq a_{n}\leqq\ldots\leqq M$.  Therefore the set  $\{a_{1},\,a_{2},\,\ldots\}$ has a finite supremum $s\leqq M$.  We show that

 $\displaystyle\lim_{n\to\infty}a_{n}\;=\;s.$ (1)

Let $\varepsilon$ an arbitrary positive number.  According to the definition of supremum we have  $a_{n}\leqq s$  for all $n$ and on the other hand, there exists a member $a_{n(\varepsilon)}$ of the sequence that is $>s-\varepsilon$.  Then we have  $s-\varepsilon,  and since the sequence is nondecreasing,

 $0\;\leqq\;s-a_{n}\;\leqq\;s\!-\!a_{n(\varepsilon)}\;<\;\varepsilon\quad\mbox{% for all}\;\,n\geqq n(\varepsilon).$

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 $L$ converges to a limit which is not less than $L$.

Note.  A good application of the latter theorem is in the proof that Euler’s constant exists.

 Title limit of nondecreasing sequence Canonical name LimitOfNondecreasingSequence Date of creation 2013-03-22 17:40:31 Last modified on 2013-03-22 17:40:31 Owner pahio (2872) Last modified by pahio (2872) Numerical id 13 Author pahio (2872) Entry type Theorem Classification msc 40-00 Synonym nondecreasing sequence with upper bound Synonym limit of increasing sequence Related topic MonotonicallyIncreasing Related topic MonotoneIncreasing Related topic Supremum Related topic Infimum Related topic ConvergenceOfTheSequence11nn