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Homelimit of nondecreasing sequence

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# limit of nondecreasing sequence

Theorem. A monotonically nondecreasing sequence of real numbers with upper bound a number $M$ converges to a limit which does not exceed $M$.

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 $s-\varepsilon<a_{{n(\varepsilon)}}\leqq s$, 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 $L$.

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## Comments

## Nondecreasing sequence with upper bound

I did not find this important theorem in PM, and therefore I wrote this entry. It were good if this entry had alternative titles -- for that one could more easily find it. Can someone propose alternative titles?

Jussi

## Re: Nondecreasing sequence with upper bound

Why not just call it limit of nondecreasing sequence? I realize that leaves out the "upper bound" part, but it includes the most important things that people will likely search for.

Roger

## Re: Nondecreasing sequence with upper bound

Thanks, Roger, you may be right. I change the title.

Jussi

## Re: Nondecreasing sequence with upper bound

In Portuguese everybody calls it: "bounded monotonic sequences are convergent".

But I like the current title..

Cheers,

Rui