# Does non-decreasing sequence of this form converge?

Given a non-decreasing sequence $(a_n)$: $$a_1 \leq a_2 \leq a_3 \leq a_4 \ldots$$ and $$\displaystyle\lim_{n\to\infty}(a_n - a_{n-1}) = 0$$ Does it have to converge?
For strictly than sequence $a_1 < a_2 < a_3 < a_4 < \ldots$ with the limit property, it's easy to show that it doesn't converge, for example take $a_n = \sqrt{n}$. In this case, however I couldn't find a counter example sequence, and I have a feeling this sequence might converge but again I'm not so sure. Any hint would be greatly appreciated.

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$a_n = \sqrt{n}$ also satisfies $a_1 \leq a_2 \leq a_3 \leq a_4 \ldots$. So too does $a_n = \sqrt{\lfloor n/2\rfloor}$ –  Henry Sep 30 '12 at 9:02
As Henry indicates, you seem to have answered this yourself already. A strictly increasing sequence is in particular a non-decreasing sequence, so. . . –  saurs Sep 30 '12 at 13:00
@Henry: That was a clever fix, I was thinking of $n \pmod {2}$. Thanks. –  Chan Sep 30 '12 at 19:05

Clearly the the sequence $b_n=a_{n+1}-a_n$ is non-negative, i.e. $b_n\ge0$ for each $n$.

• If any non-negative sequence $b_n\ge0$ is given, can you find a corresponding (non-decreasing) sequence $a_n$ such that $b_n=a_{n+1}-a_n$?
• Can you find a non-negative sequence which does not have limit?
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Great hint! Thanks a lot ;) –  Chan Sep 30 '12 at 6:58
Take for example the harmonic sequence: $$H_n = 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$
It has the property that $H_n \to \infty$, but $H_{n}-H_{n-1}=\frac{1}{n} \to 0$.