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

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?

An easy way to visualize why this can't be true is to try putting some points on a number line.

Start with 1 point in [0, 1):

2 points in [1, 2):

And so on:

Now you have a sequence that grows to infinity but keeps getting closer together.

No. Just consider the case in which $a_n=1+\frac12+\frac13+\cdots+\frac1n$. Note that then we have$$\lim_{n\to\infty}a_{n+1}-a_n=\lim_{n\to\infty}\frac1{n+1}=0.$$But $a_n$ is the $n$^th partial sum of the harmonic series, and therefore $(a_n)_{n\in\Bbb N}$ diverges.

Any increasing sequence $\{a_n\}_{n\geq 1}$ has limit in $\mathbb{R}\cup\{+\infty\}$. It is $\sup_{n\geq 1} a_n$. Such $\sup$ or supremum can be a finite number or $+\infty$ (even if we know that $a_{n+1}-a_n\to 0$).

An example with a finite limit is $a_n=1-1/n\to 1$ and $a_{n+1}-a_n=\frac{1}{n(n+1)}\to 0$.

On the other hand $a_n=\sqrt{n}\to +\infty$ and $a_{n+1}-a_n=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.

So, the answer is NO, the condition $a_{n+1}-a_n\to 0$ is not sufficient for an increasing sequence $\{a_n\}_{n\geq 1}$ to have a FINITE limit.

Another counterexample is $a_n=\ln n$, for $n\geq1$. The difference of successive terms is $\ln(n+1)-\ln n = \ln (1+1/n) \rightarrow \ln 1 = 0$, as $n \rightarrow \infty$, yet $\ln n$ itself tends to infinity, as $n$ tends to infinity.