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

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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).

Answer 7

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.