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?

  • 6
    $\begingroup$ Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question. $\endgroup$
    – Adayah
    Commented Feb 17, 2019 at 18:26
  • 15
    $\begingroup$ Have you tried logarithms? $\endgroup$
    – user541686
    Commented Feb 18, 2019 at 6:14
  • 5
    $\begingroup$ Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge. $\endgroup$
    – Bar Alon
    Commented Feb 18, 2019 at 12:44
  • $\begingroup$ The harmonic series should answer this question for you $\endgroup$
    – MPW
    Commented Feb 19, 2019 at 21:56
  • 1
    $\begingroup$ No it converges iff the difference of consecutive terms forms a summable series, which is stronger than just converging to zero. $\endgroup$
    – shalop
    Commented Feb 20, 2019 at 2:29

7 Answers 7


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):

number line showing single point at 0

2 points in [1, 2):

number line showing points at 0, 1, 1.5

And so on:

number line showing points at 0, 1, 1.5, 2, 2.33, 2.67, 3, 3.25, ...

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

  • 24
    $\begingroup$ +1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges. $\endgroup$ Commented Feb 18, 2019 at 0:21
  • 10
    $\begingroup$ This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not $\endgroup$
    – gota
    Commented Feb 18, 2019 at 12:33
  • 11
    $\begingroup$ Note that this is (approximately) the same as the sequence $a_n=\sqrt{n}$. $\endgroup$
    – tomasz
    Commented Feb 18, 2019 at 12:41
  • 6
    $\begingroup$ The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence. $\endgroup$ Commented Feb 18, 2019 at 14:07
  • $\begingroup$ @MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index. $\endgroup$
    – Owen
    Commented Feb 18, 2019 at 22:49

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.

  • 9
    $\begingroup$ Rhys: His sequence is $1, \frac32, \frac{11}6, \frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum. $\endgroup$
    – M D
    Commented Feb 17, 2019 at 16:49
  • 5
    $\begingroup$ @RhysHughes $a_n=1+\frac12+\cdots+\frac1n$ IS increasing and $a_{n+1}-a_n=\frac{1}{n+1}\to 0$. $\endgroup$
    – Robert Z
    Commented Feb 17, 2019 at 16:49
  • $\begingroup$ Isn't $a_{n+1} - a_n = \frac 1{n+1} - \frac 1n = \frac n{n(n+1)} - \frac{n+1}{n(n+1)} = \frac {n-(n+1)}{n(n+1)} = \frac {-1}{n^2+n} = -\frac 1{n^2+n} \to 0$? $\endgroup$
    – wizzwizz4
    Commented Feb 18, 2019 at 20:28
  • 2
    $\begingroup$ @wizzwizz4 No, because $a_n \neq \frac{1}{n}$. Instead, $a_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n}$. So, $$a_{n+1} - a_n = \left(1 + \frac{1}{2} + \ldots + \frac{1}{n} + \frac{1}{n+1}\right) - \left(1 + \frac{1}{2} + \ldots + \frac{1}{n}\right) = \frac{1}{n+1}.$$ $\endgroup$ Commented Feb 18, 2019 at 22:39
  • $\begingroup$ @MD: a series is a sequence. $\endgroup$
    – Taladris
    Commented Feb 19, 2019 at 4:15

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.

  • 2
    $\begingroup$ Which is, in a way, the same counterexample, because $\sum_{k=1}^n\frac1k = \ln n + \gamma + \mathcal O\left(\frac1n\right)$. $\endgroup$ Commented Feb 18, 2019 at 20:43

No. Consider the sequence $\{a_n\}_{n=1}^\infty$ given by

  • $a_n = \sum\limits_{k=1}^{n} \frac{1}{k}$.

It follows that

  • $a_n > a_{n-1}$
  • $a_n - a_{n-1} = \frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$, but
  • $a_n = \sum\limits_{k=1}^{n} \frac{1}{k} \rightarrow \infty$ as $n \rightarrow \infty$ (by, e.g., integral test).

Note that if we define $b_n=a_{n+1}-a_n$, then $a_n=a_0+\sum_{n=0}^{\infty}b_n$. So this question is equivalent to asking whether the terms of an infinite series going to zero is sufficient for the series to converge. There are a variety of examples of series with terms that go to zero, yet do not converge, with the harmonic series ($\sum \frac 1 n$) being one of the most famous.

And in fact we can construct a counterexample from any sequence by defining a sequence $c_n$ by simply re-indexing the terms. We set $c_0$ equal to $a_0$. Then set $c_{k1}$ equal to $a_1$, where $k_1>a_1-a_0$, and fill in the terms $c_1$ to $c_{k-1}$ with equally spaced terms; this will result in all of the consecutive differences from $c_0$ to $c_{k1}$ being less than $1$. Then set $c_k2$ equal to $a_2$, where $k_2+k_1>2(a_2-a_1)$, which results in consecutive differences between $c_{k1}$ to $c_{k2}$ being less than $\frac 1 2$. Just keep re-indexing each term and filling in more and more new terms, and you can drive the consecutive differences arbitrarily low.

Another approach is to look at a sequence as an approximation of a continuous function, and the difference between successive terms as an approximation of the derivative. Then we just need a function such that $f'(x)$ converges to zero, but $f$ diverges. Two examples of such are the log function, (which gives a sequence very similar to the harmonic sequence) and square root. Note that both of these examples can be obtained by taking the inverse of a function whose derivative is constantly increasing. If $g'$ goes to infinity, then $(g^{-1})'$ goes to zero. But if the domain of $g$ is the whole real line, then the range of $g^{-1}$ is the whole real line, i.e. $g^{-1}$ goes to infinity.


The condition $a_{n+1}-a_n \to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $\lim\limits_{n\to\infty}(a_{n+m(n)}-a_n)=0$ for all $m(n)\in \mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.

Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.

  • 3
    $\begingroup$ Your stated condition $\lim_{n \to \infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = \log n$. I think what you would want is that $\lim_{n \to \infty} (a_{n+m}-a_n) = 0$ uniformly in $m$. $\endgroup$ Commented Feb 17, 2019 at 17:05
  • $\begingroup$ @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it? $\endgroup$ Commented Feb 17, 2019 at 17:09
  • $\begingroup$ Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice. $\endgroup$ Commented Feb 17, 2019 at 17:22
  • $\begingroup$ @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused. $\endgroup$ Commented Feb 17, 2019 at 17:25
  • $\begingroup$ I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct? $\endgroup$
    – supercat
    Commented Feb 18, 2019 at 19:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .