Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 8525

Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?

$
0
0

If $a_n$ is a sequence such that$$a_1 \leq a_2 \leq a_3 \leq \dotsb$$ and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?

I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing.

$$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$

This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.

So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?


Viewing all articles
Browse latest Browse all 8525

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>