# 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?

• 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. – Adayah Feb 17 '19 at 18:26
• Have you tried logarithms? – user541686 Feb 18 '19 at 6:14
• 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. – Bar Alon Feb 18 '19 at 12:44
• The harmonic series should answer this question for you – MPW Feb 19 '19 at 21:56
• No it converges iff the difference of consecutive terms forms a summable series, which is stronger than just converging to zero. – shalop Feb 20 '19 at 2:29

## 7 Answers

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.

• 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. – M D Feb 17 '19 at 16:49
• @RhysHughes $a_n=1+\frac12+\cdots+\frac1n$ IS increasing and $a_{n+1}-a_n=\frac{1}{n+1}\to 0$. – Robert Z Feb 17 '19 at 16:49
• Isn't $a_{n+1} - a_n = \frac 1{n+1} - \frac 1n = \frac n{n(n+1)} - \frac{n+1}{n(n+1)} = \frac {n-(n+1)}{n(n+1)} = \frac {-1}{n^2+n} = -\frac 1{n^2+n} \to 0$? – wizzwizz4 Feb 18 '19 at 20:28
• @wizzwizz4 No, because $a_n \neq \frac{1}{n}$. Instead, $a_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n}$. So, $$a_{n+1} - a_n = \left(1 + \frac{1}{2} + \ldots + \frac{1}{n} + \frac{1}{n+1}\right) - \left(1 + \frac{1}{2} + \ldots + \frac{1}{n}\right) = \frac{1}{n+1}.$$ – Theo Bendit Feb 18 '19 at 22:39
• @MD: a series is a sequence. – Taladris Feb 19 '19 at 4:15

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.

• +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. – Theo Bendit Feb 18 '19 at 0:21
• This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not – gota Feb 18 '19 at 12:33
• Note that this is (approximately) the same as the sequence $a_n=\sqrt{n}$. – tomasz Feb 18 '19 at 12:41
• 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. – Marc van Leeuwen Feb 18 '19 at 14:07
• @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. – Owen Feb 18 '19 at 22:49

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.

• Which is, in a way, the same counterexample, because $\sum_{k=1}^n\frac1k = \ln n + \gamma + \mathcal O\left(\frac1n\right)$. – Roman Odaisky Feb 18 '19 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.

• 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 $m$. – Nate Eldredge Feb 17 '19 at 17:05
• @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it? – Haris Gušić Feb 17 '19 at 17:09
• Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice. – Nate Eldredge Feb 17 '19 at 17:22
• @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused. – Haris Gušić Feb 17 '19 at 17:25
• 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? – supercat Feb 18 '19 at 19:27