When does $\lim\limits_{n\to\infty}|a_{n+1}-a_n|=0$ imply $a_n$ convergent?

Question

Is it true: Let $\{a_n\}$ be real sequence. If $|a_{n+1}-a_n|<\frac1{3^n}$ for all $n$, then $\{a_n\}$ convergent.

I am asking this, because I was pointed out that there is a real sequence $\{a_n\}$ such that $|a_{n+1}-a_n|<\frac1{n}$, yet $\{a_n\}$ not convergent.

Answer 1

Note that the sequence $(a_n)_n$ converges if, and only if, the series $\sum_n (a_{n+1}-a_n)$ converges. (Can you see why?)

But if we have $$\forall n\,\qquad \lvert a_{n+1}-a_n \rvert < \frac{1}{3^n}$$ then by comparison with $\sum_n \frac{1}{3^n}$ the series $\sum_n \lvert a_{n+1}-a_n \rvert$ converges, i.e., the series $\sum_n (a_{n+1}-a_n)$ converges absolutely. And therefore converges.

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Now, the thing in this argument which fails when we are only told $\lvert a_{n+1}-a_n \rvert < \frac{1}{n}$ is when comparing to the series $\sum_n \frac{1}{n}$... as this series does not converge.

Answer 2

Let $m > n$. Now,

$$ |a_m - a_n| \leq \sum_{i=n}^{m+1}|a_{i+1} - a_i| \leq \sum_{i=n}^{m+1}\frac{1}{3^i} \leq \sum_{i \geq n}\frac{1}{3^i} \xrightarrow{n \to \infty} 0 $$

Hence $(a_n)_{n \geq 1}$ is a Cauchy sequence and therefore by completeness of the reals, it converges. As someone else has said, we can even give a equivalency between the convergence of $(a_n)_{n \geq 1}$ and the convergence of $\sum_{i \geq i}a_{i+1} - a_i$. Note that this may fail for $\frac{1}{n}$ because it is not summable: as an explicit example, if

$$ s_n = \sum_{i = 1}^n\frac{1}{i} $$

then $(s_n)_{n \geq 1}$ does not converge but $|s_{n+1} -s_n| = \frac{1}{n+1} < \frac{1}{n}$ for each $n \geq 1$.

Answer 3

Hint: if $|a_{n+1}-a_n|<\frac{1}{3^n}$, how large can $\sum_{k=0}^\infty|a_{n+k+1}-a_{n+k}|$ be? What does this tell you about $|a_m-a_n|$ if $m>n$?

Answer 4

One sequence $a_n$ which diverges even though $|a_{n+1}-a_n|\to 0$ is $$ a_n=\sum_{k=1}^n\frac1n\\ a_{n+1}-a_n=\frac1{n+1} $$ The general requirement for convergence by comparing the terms of the sequence (and not referring to an actual limit which may or may not be one of the terms of the sequence), is the Cauchy criterion:

A sequence $b_n$ is Cauchy iff for any $\varepsilon>0$, there is an $N\in \Bbb N$ such that for any $m,n>N$ we have $|a_m-a_n|<\varepsilon$.

In other words, not only do we have a requirement that $|b_{n+1}-b_n|\to0$, but also $$ |b_{n+2}-b_n|,|b_{n+3}-b_n|,\ldots $$ should all go to zero "together", in the sense that the set $$ \{|b_{n+1}-b_n|,|b_{n+2}-b_n|,|b_{n+3}-b_n|,\ldots\} $$ Is bounded, and that bound goes to $0$ as $n\to\infty$.

In more general spaces, where a sequence might look like it "ought" to converge, but it doesn't because an actual limit doesn't exist in the space (the set of rational numbers is one example), Cauchy is the formal criterion for "ought to converge".

Answer 5

In general is the series of general term $s_n = |a_{n+1} -a_n|$ converges then the sequence $(a_n)_n$ converges too.

Indeed, assume $\sum s_n<\infty$ then $(a_n)_n$ will a Cauchy sequence as follows , $$ |a_{n+p} -a_n| \le \sum_{j=0}^{p-1} |a_{n+j+1}-a_{n+j}|= \sum_{j=0}^{p-1} s_{n+j} = \sum_{k= n}^{n+p-1} s_{k} \le \sum_{k= n}^{\infty} s_{k} \to 0\qquad \hbox{as}~~n\to \infty.$$

However we have some counter example . Take $a_n = \ln(n)$ or $a_n = \sqrt{n}$

Check that in both cases, $$|a_{n+1}-a_n| \to 0$$

Whereas, $(a_n)_n$ heavily diverges.